Wow, this problem is cool. Let's create a random subset of $[n]$ by including each $k=1,2,\ldots,n$ independently, with probability $x\in(0,1).$ Then, $f(x)$ denotes the probability that the set we've created belongs to $\mathcal{A}.$ Notice that because $\mathcal{A}$ is an upset, then there exists a collection of sets $K_1,\ldots,K_m$ with the property that $S\in\mathcal{A}$ if and only if $K_i\subseteq S$ for some $i=1,\ldots,m.$ Consequently, letting $E_i$ be the event that $K_i\subseteq S$ and given that $P(K\subseteq S)=x^{|K|}$ it follows by PIE that \[f(x)=P\left(\bigcup_{i=1}^m E_i\right)=\sum_{t=1}^m\sum_{i_1,\ldots,i_t} P\left(\bigcap_{j=1}^tE_{i_j}\right)(-1)^{t+1}=\sum_{t=1}^m\sum_{i_1,\ldots,i_t}x^{|K_{i_1}\cap\hspace{1pt}\cdots\hspace{1pt}\cap\hspace{1pt}K_{i_t}|}(-1)^{t+1}.\]To show that $f(x)$ is increasing, it suffices to show that $f'(x)\geqslant 0$ which is equivalent to \[\sum_{t=1}^m\sum_{i_1,\ldots,i_t}|K_{i_1}\cap\cdots\cap K_{i_t}|\cdot x^{|K_{i_1}\cap\hspace{1pt}\cdots\hspace{1pt}\cap\hspace{1pt}K_{i_t}|}(-1)^{t+1}\geqslant 0.\]Now, the key is to reinterpret $|K_{i_1}\cap\cdots\cap K_{i_t}|.$ For each $k=1,\ldots,n$ let $\mathcal{A}_k$ be the set of indices $i{}$ for which $k\in K_i.$ Let $\chi_S(x)$ be the characteristic function of $x\in S.{}$ Of course, $k\in K_{i_1}\cap\cdots\cap K_{i_t}$ if and only if $K_{i_1},\ldots,K_{i_t}\in \mathcal{A}_k.$ Thus\begin{align*}f'(x)&=\sum_{t=1}^m\sum_{i_1,\ldots,i_t}\Bigg(\sum_{i=1}^n\chi_i(K_{i_1}\cap\cdots\cap K_{i_t})\Bigg) x^{|K_{i_1}\cap\hspace{1pt}\cdots\hspace{1pt}\cap\hspace{1pt}K_{i_t}|}(-1)^{t+1} \\ &=\sum_{i=1}^n\Bigg(\sum_{t=1}^{|\mathcal{A}_i|}\sum_{\substack{i_1,\ldots,i_t \\ i_j\in\mathcal{A}_i}}x^{|K_{i_1}\cap\hspace{1pt}\cdots\hspace{1pt}\cap\hspace{1pt}K_{i_t}|}(-1)^{t+1}\Bigg)=\sum_{i=1}^n P\left(\bigcup_{j\in\mathcal{A}_i}E_j\right)\geqslant 0.\end{align*}Remarks. This looks approachable using the Ahlswede-Daykin theorem or perhaps the Fortuin-Kasteleyn-Ginibre inequality. I'm really curious to see if there are any solution involving these reuslts, or at least motivational arguments.

Isn't it much easier? If we increase the probability $x$, we can interpret this as first choosing the element with probability $x$, and then choosing the element (if not yet chosen) with another given (conditional) probability. But realizing both stochastic experiments in this way on the same probability space, we always get a superset of the set with the smaller probability for the larger probability. But since our family contains all supersets of its elements, the conclusion now follows immediately.

Fun fact, me and the proposer knew a nice non-olympiad original source of this problem, but I just unexpectedly heard about the coincidence with VJIMC 2007 Category 2 Problem 4