Group generated by a proper subgroup and an element of order $2$ has index $2$

Since $G$ is generated by $H$ and $G$, and $wHw^{-1} = H$, we can show that any element in the group is either an element of $H$, or it is of the form $hw$ for some $h\in H$. This means that $H$ has two cosets $H$ and $Hw$, and therefore $[G:H] = 2$.

I made a claim above. Let's show that. Take an element in $G$. Since $G$ is generated by $H$ and $w$, it can be written as some finite product on (at least one) one of the following four forms: $$ wh_1wh_2wh_3\cdots h_{n-1}wh_nw\\ wh_1wh_2wh_3\cdots h_{n-1}wh_n\\ h_1wh_2wh_3\cdots h_{n-1}wh_nw\\ h_1wh_2wh_3\cdots h_{n-1}wh_n $$ Consider what $wHw^{-1} = H$ means. It means that for any $h\in H$, there is a $h'\in H$ such that $$ whw^{-1} = h'\\ wh = h'w $$ Which is to say, any time in that product where we have a $w$ to the left of a $h_k$, we can move that $w$ to the other side of the $h_k$, as long as we change that $h_k$ to some corresponding $h'_k$.

This is something we can keep doing until all the $w$'s are to the right, and all the $h_i$'s are on the left, so our product becomes $$ h_1'h_2'\cdots h_n'ww\cdots w $$ That long product of $h_i'$'s results in a single element $h\in H$. And we know that $w$ has order $2$, so that long product of $w$'s is either going to end up being the identity, or just $w$. Which is to say, our arbitrary element of $G$ is of either the form $h$, or the form $hw$.