Let $G$ be a finite group and let $M,N \lhd G$ be normal subgroups with a trivial intersection. Suppose that $G$ has a subgroup of index $2$. Must $G$ have a subgroup of index $2$ which contains either $M$ or $N$?

Then answer is, in general, no. Let $G = {\rm SL}(2,5) \times M$, where $M$ is cyclic of order $2$. Then $V = Z(G)$ is a Klein $4$-group. Let $H = {\rm SL}(2,5)$. Let $N$ be a subgroup of order $2$ of $V$ with $M \neq N$ and $N \not \leq H.$ There is such a subgroup, as $V$ has three subgroups of order $2$. Now we have $H = G^{\prime}$ as $H = H^{\prime}.$ Thus $H$ is the unique subgroup of $G$ of index $2$ ( for if $[G:K] =2,$ then $K \lhd G$ and $G/K$ is Abelian, so $G^{\prime} \leq K$). Hence in this case, we have $M \lhd G,N \lhd G,M \cap N = 1$, but no subgroup of index $2$ of $G$ contains either of $M$ or $N.$