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Consider the following proposition:

Proposition. Let $H$ and $K$ be finite index subgroups of a group $G$. Then $H\cap K$ is also a finite index subgroup of $G$.

Proof. Let $X$ be a path connected $2$-complex with a chosen point $x_0$ such that $\pi_1(X, x_0)\cong G$. Since $X$ admits a universal cover, we can find connected coverings $p:A\to X$ and $q:B\to Y$ with $p_*(\pi_1(A, a_0))=H$ and $q_*(\pi_1(B, b_0))=K$ for some points $a_0\in A$ and $b_0\in B$.

Note. For a cover $\phi:Y\to X$, the stabilizer of a point $y\in \phi^{-1}(x_0)$ under the $\pi_1(X, x_0)$ action is same as $\phi_*(\pi_1(Y, y))$.

Define an action of $\pi_1(X, x_0)$ on $p^{-1}(x_0)\times q^{-1}(x_0)$ given by $[\gamma]\cdot(a, b)=([\gamma]\cdot a, [\gamma]\cdot b)$. Now $H$ and $K$ are the stabilizers of $a_0\in p^{-1}(x_0)$ and $b_0\in q^{-1}(x_0)$ under the respective $\pi_1$-action. Thus the stabilizer of $(a_0, b_0)\in p^{-1}(x_0)\times q^{-1}(x_0)$ is $H\cap K$. Since $p^{-1}(x_0)\times q^{-1}(x_0)$ is finite, so is the the index of $H\cap K$ in $\pi_1(X, x_0)$.

This finishes the proof.

So basically what covering spaces help us do here is the following: For a given finite index subgroup $S$ of $G$, we are able to construct an action of $G$ on a finite set $F$ which has a point $p$ such that the stabilizer of $p$ is $S$. From here algebra takes over. Is everything alright? Thanks.

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Using covering spaces in this manner seems like overkill.

For any subgroup $S$ of $G$, the left multiplication action of $G$ on itself preserves the decomposition of $G$ into left cosets of $S$, because $g(aS)=(ga)S$. So you can simply take $F$ to be the set of left cosets of $S$ in $G$. The left coset $S \in F$ is a point of $F$ whose stabilizer is $S$.

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    Right! I had another argument in mind using Pigeohole. If $[G:H]=m$ and $[G:K]=n$, then we show that $[G:H\cap K]\leq mn$. For if $g_0H\cap K, g_1H\cap K, \ldots, g_{mn}H\cap K$ are $mn+1$ cosets of $H\cap K$, then by PHP, some $n+1$ of the $g_i$'s, call them $g_{i_0}, \ldots, g_{i_n}$, fall in the same coset of $H$. Again by PHP, some two of the $g_{i_j}$'s, call them $a$ and $b$, fall in the same coset of $K$. Thus $aH\cap K=bH\cap K$,where $a=g_k$ and $b=g_\ell$ for some $k\neq \ell$. Thus the cosets $g_iH\cap K$'s are not pairwise distinct and we are done. – 2017-02-04