Does this work? Firstly, we use similar expression for winding number in the problem statement, i.e., $\deg_O{\mathcal{L}}$ is the winding number for a polygonal line $\mathcal{L}$ around $O$. Let $P_i$ denote a self-intersection of the sides of $\mathcal{L}$. Also, $\mathbb{P}$ denote the set of all such $P_i$'s. Now we induct upon $|\mathbb{P}|$. Firstly the base case, $|\mathbb{P}|=0$, i.e., when the polygonal line has no self-intersections. $\textbf{CLAIM: }$When $|\mathbb{P}|=0$, then $\deg_O{\mathcal{L}}=1$. $\textbf{PROOF: }$We induct on the number of vertices. Note that $\mathcal{L}$ is symmetric with respect to $O$, so $O$ lies in the interior of $\mathcal{L}$. So the number of vertices increases by $2$ on adding another vertex due to symmetry. The case for number of vertices $=4$ is trivial. Now consider the induction hypothesis for $n$-gon $\mathcal{L}'$ with $n$ vertices. We prove our case for the $(n+2)-$gon $\mathcal{L}$. Select any arbitrary vertex $A_i$ of the $(n+2)-$gon. $A_i'$ denote the symmetric vertex of $A_i$ wrt $O$. Thus, \begin{align*} \deg_O{\mathcal{L}}&=\dfrac{\angle A_1OA_2+\cdots+\angle A_{i-1}OA_{i}+\angle A_{i}OA_{i+1}+\cdots+\angle A_{i-1}'OA_{i}'+\angle A_{i}'OA_{i+1}'+\cdots+\angle A_nOA_1}{2\pi}\\ &=\dfrac{\angle A_1OA_2+\cdots+\angle A_{i-1}OA_{i+1}+\cdots+\angle A_{i-1}'OA_{i+1}'+\cdots+\angle A_nOA_1}{2\pi}\\ &=\deg_O{\mathcal{L}'}\\ &=1. \end{align*} Now as $\mathcal{L}$ is symmetric with respect to $O$, we have that $|\mathbb{P}|$ increases by $2$ (since if $P_1$ becomes a point of self intersection, then it's reflection $P_2$ wrt $O$ also becomes a point of self-intersection). So, let $P_1,P_2$ be two such symmetric points of self intersections, wrt $O$. We firstly have the following lemma. $\textbf{LEMMA: }$If $P$ is a point of self-intersection of $\mathcal{L}$, and $\mathcal{L}_1$ and $\mathcal{L}_2$ denote the polygonal lines formed by breaking $\mathcal{L}$ at $P$. Then $\deg_O{\mathcal{L}}=\deg_O{\mathcal{L}_1}+\deg_O{\mathcal{L}_2}$. $\textbf{PROOF: }$Firstly, $P$ denote the intersection of $A_iA_{i+1}$ and $A_jA_{j+1}$ where $j>i$. We use oriented angles for the following angel chase. Note that, \begin{align*} \deg_O{\mathcal{L}}&=\dfrac{\angle A_1OA_2+\cdots+\angle A_iOA_{i+1}+\cdots+\angle A_jOA_{j+1}+\cdots+\angle A_nOA_1}{2\pi}\\ &=\dfrac{\angle A_1OA_2+\cdots+\angle A_iOP+\angle POA_{i+1}+\cdots+\angle A_jOP+\angle POA_{j+1}+\cdots+\angle A_nOA_1}{2\pi}\\ &=\tfrac{(\angle A_1OA_2+\cdots+\angle A_iOP+\angle POA_{j+1}+\cdots+\angle A_nOA_1)+(\angle POA_{i+1}+\angle A_{i+1}OA_{i+2}+\cdots+\angle A_{j-1}OA_{j}+\angle A_jOP)}{2\pi}\\ &=\tfrac{(\angle A_1OA_2+\cdots+\angle A_iOP+\angle POA_{j+1}+\cdots+\angle A_nOA_1)}{2\pi}+\tfrac{(\angle POA_{i+1}+\angle A_{i+1}OA_{i+2}+\cdots+\angle A_{j-1}OA_{j}+\angle A_jOP)}{2\pi}\\ &=\deg_O{\mathcal{L}_1}+\deg_O{\mathcal{L}_2}. \end{align*} Thus we now have $\deg_O{\mathcal{L}}=\deg_O{\mathcal{L}_1}+\deg_O{\mathcal{L}_2}$. Notice as $\mathcal{L}$ is symmetric wrt $O$, we can perform a similar breaking at $P_2$ to form a $\mathcal{L}_1'$ from $\mathcal{L}_2$, where $\mathcal{L}_1'$ is the reflection of $\mathcal{L}_1$ over $O$. Notice that we can't actually perform this breaking if $\mathcal{L}_1$ and $\mathcal{L}_1'$ are the same polygons, i.e., if $\mathcal{L}_1$ is symmetric wrt $O$. Thus we have the following Claim to avoid this. Also, $\mathcal{L}_1$ being symmetric wrt $O\implies P_2\in\mathcal{L}_1$. $\textbf{CLAIM: }$If $P_1$ be a point of self-intersection of $\mathcal{L}$. $\mathcal{L}_1$ and $\mathcal{L}_2$ denote the polygons as defined previously. Then at least one of $\mathcal{L}_1$ or $\mathcal{L}_2$ is non-symmetric wrt $O$. $\textbf{PROOF: }$FTSOC, assume both $\mathcal{L}_1$ and $\mathcal{L}_2$ are symmetric wrt $O$. WLOG, let the orientation of the side $A_iA_{i+1}$ of be as such shown in the diagram. Note that the addition of $P_1,P_2$ does not affect the orientation of the edges. Now note that due to the orientation of $A_iA_{i+1}$ as shown by the triple-arrowed edge in the figure, it determines the orientation of the next edges too as shown by the single-arrowed blue sides, and moreover, the orientation of the edge goes unchanged through $P_1,P_2$. Thus we finally are able to form a cycle as shown by the blue coloured polygon. Similarly, we can orient the edges the other half marked in red in a similar manner forming another cycle. Thus, if we temporarily remove the points $P_1,P_2$, we are able to form two disjoint cycles of $\mathcal{L}_1$ and $\mathcal{L}_2$, but since $\mathcal{L}$ is a polygonal line, it cannot be broken into disjoint cycles, and thus we arrive a contradiction. Now WLOG, assume that $\mathcal{L}_1$ is not symmetric wrt $O$. $\mathcal{L}_1'$ denote the reflection of $\mathcal{L}_1$ wrt $O$. Thus we can finally apply another breaking at $P_2$, thus resulting in the formation of $\mathcal{L}_1'$ from $\mathcal{L}_2$ and let the remaining polygon be named $\mathcal{L}_0$. Also, as we are removing $\{\mathcal{L}_1,\mathcal{L}_1'\}$ which are symmetric to each other wrt $O$, and $\mathcal{L}$ itself is symmetric wrt $O$, we get that $\mathcal{L}_0$ is also symmetric wrt $O$. Thus we have currently at hand three distinct polygons, $\mathcal{L}_1,\mathcal{L}_1',\mathcal{L}_0$, where $\{\mathcal{L}_1,\mathcal{L}_1'\}$ are symmetric to each other wrt $O$. This means that $\deg_O{\mathcal{L}_1}=\deg_O{\mathcal{L}_1'}$. Therefore, we finally have $\deg_O{\mathcal{L}}=\deg_O{\mathcal{L}_1}+\deg_O{\mathcal{L}_1'}+\deg_O{\mathcal{L}_0}=2\deg_O{\mathcal{L}_1}+\deg_O{\mathcal{L}_0}$. Now due to the induction we performed earlier, we now have that $\deg_O{\mathcal{L}_0}$ is $\operatorname{odd}$. So, $\deg_O{\mathcal{L}}=2\deg_O{\mathcal{L}_1}+\deg_O{\mathcal{L}_0}=\operatorname{even}+\operatorname{odd}=\operatorname{odd}$ and we are done.

Woah winding numbers! This problem seems interesting. Can I solve it?

Complex Analysis FTW. Anyways, let $\mathcal{\chi}(\mathcal L(t))$ denote the winding number of the polygonal line $\mathcal L(t): [-1,1]\rightarrow \mathbb C$ WRT $O$. Deform $\mathcal L$ in a sufficiently small neighborhood around the vertices so that no new intersections are formed, $\mathcal L$ is still symmetric and $\mathcal L(t)$ becomes differentiable. Ensure that $\mathcal L(-1)$ is not an intersection point of the polygonal line. Due to the symmetry criteria, $\mathcal L(t)=-\mathcal L(1+t)$. It is well known that $\chi=\frac{1}{2\pi i}\int_{\mathcal L}d\theta=\frac{1}{2\pi i}\int_{-1}^{1}\frac{\mathcal L'(t)}{\mathcal L(t)}dt$. Define $f(x)=\int_{-1}^{x}\frac{\mathcal L'(t)}{\mathcal L(t)}dt$. Then $f'(x) = \frac{\mathcal L'(x)}{\mathcal L(x)}$ and as result $\frac{d}{dt}e^{-f(x)}\mathcal L(x)=0\implies e^{-f(x)}\mathcal L(x)$is constant. Hence, $e^{-f(x)}=-e^{-f(1+x)}$ or in other words, $f(1)=(2n+1)2\pi i$ or $\chi=2n+1$. $\blacksquare$ I spent hours on this, trying to use elementary methods. However, after I tried using some advanced results (see: https://math.stackexchange.com/questions/1197670/winding-number-demonstration) I was able to finish this in 15 mins.

It suffices to show that any odd map $S^1 \to S^1$ has odd degree. It turns out this is true in general for any $S^n$ (but I don't think this link proves it for $n=1$). You can find a simple proof for just $n=1$ using basic covering theory here. This is essentially what @above's solution is, but without analysis phrasing.