Let $ABCD$ be a parallelogram which contains our polygon $\mathcal P$ and which has minimal area (by a compactness argument we can prove that such a parallelogram does indeed exist). Let $PQ$ be the segment along which $\mathcal P$ intersects $AB$, with $Q$ between $P$ and $B$ ($P$ and $Q$ can coincide), and let $RS$ be the segment along which $\mathcal P$ intersects $CD$, with $S$ between $R$ and $D$. Suppose the line $\ell_Q$ through $Q$ parallel to $AD$ cuts $CD$ strictly between $R$ and $D$. Then keep the lines $AD,BC$ fixed, and slightly rotate $AB$ around $Q$ such that $B$ moves into the segment $BC$. At the same time, rotate $CD$ around $R$ so as to keep it parallel to $AB$. Then the distance between the lines $AD,BC$ remains fixed, but the sides on $AD,BC$ of the new parallelogram $ABCD$ are shorter, contradicting the minimality of the area of the initial parallelogram. The same happens if we assume that the line $\ell_P$ through $P$ parallel to $AD$ cuts $CD$ strictly between $S$ and $C$. This means that there is a line parallel to $AD$ which cuts $AB,CD$ in the points $X_{AB}\in[PQ],X_{CD}\in[RS]$. In the same way we show that there is a line parallel to $AB$ which cuts $AD,BC$ in the points $X_{DA},X_{BC}$ belonging to the polygon. The area of $\mathcal P$ is at least the area of the parallelogram $X_{AB}X_{BC}X_{CD}X_{DA}$, and this is precisely half the area of $ABCD$.

It clearly suffices to show that a parallelogram exists such that it contains the polygon, and has area twice or less the area of the polygon. If the polygon is a triangle, reflect it on the midpoint of one of its sides, join both triangles, you have a parallelogram that contains the triangle with area exactly twice the area of the triangle. If the polygon has $4$ or more sides, consider the quadrilateral with maximum area formed by four of its vertices, call it $ABCD$ (which is actually the polygon if the number of sides is $4$). Draw parallels to its diagonal $AC$ through $B$ and $D$, and parallels to its diagonal $BD$ through $A$ and $C$. This parallelogram contains completely the polygon inside it, since otherwise wlog a vertex $V$ would exist farther than $A$ from diagonal $BD$ and on the same side of $BD$, but then $VBCD$ would have area larger than $ABCD$, contradiction. The area of the parallelogram is exactly twice the area of $ABCD$, hence at most twice the area of the polygon with equality iff $ABCD$ is the polygon.