One question about proof between orientation and fundamental group Here is the statement on which I am now focusing,

Let M be a connected smooth manifold such that, for every $p∈M$, the
fundamental group $π_1(M,p)$ has no subgroup of index $2$ . Prove that
$M$ is orientable.

And I am following the Alozio Macedo's answer. The answerer assumed that $M$ is non-orientable, but I wonder the assumption affects the proof. I have yet not  found that where the assumption, non-orientation, is used.
Of course, in the first paragraph, the answerer said that : Assume $M$ is not orientable. We $\underline{then}$ have that the orientable double cover $\overset{\sim}M$ of $M$ is path-connected. Here is the very part that is likely to be applied for non-orientable assumption. However, according to the Hatcher's textbook (234 page) , the following is referred.

Every  manifold $M$ has an orientable, two-sheeted covering
space $M$ ...
(Note. I have regarded the given manifold $M$ as the topological manifold)

In other words, whether the given (topological) manifold is orientable or not, it absolutely has   orientable double cover $\overset{\sim}M$. Therefore, within the proof, I have not caught the part that such assumption is used.
 A: If $M$ is non-orientable, the orientable double cover is path-connected, while if $M$ is orientable, the orientable double cover is not path-connected (it is homeomorphic to $M\sqcup M$).
As pointed out by Paul Frost in the comment below, the path-connectedness of $\widetilde{M}$ is needed to conclude that the index of $f_{\#}(\pi_1(\widetilde{M}))$ in $\pi_1(M)$ is equal to the number of sheets of the cover. In the case that $M$ is orientable, we have $\widetilde{M} = M\sqcup M$ and $f : \widetilde{M} \to M$ is just $\operatorname{id}_M\sqcup \operatorname{id}_M$. Since $\widetilde{M}$ is not path-connected, we need to choose a path-connected component to talk about the fundamental group; in this case, the choice of component doesn't change the isomorphism type of the group. Note that $\pi_1(\widetilde{M}) \cong \pi_1(M)$ and $f_{\#} = (\operatorname{id}_M)_{\#} = \operatorname{id}_{\pi_1(M)}$, so $f_{\#}(\pi_1(\widetilde{M})) \cong \pi_1(M)$. In particular, the index of $f_{\#}(\pi_1(\widetilde{M}))$ in $\pi_1(M)$ is one, which is not equal to the number of sheets in the cover $f : \widetilde{M} \to M$.
