Would just like a sanity check.
I don't see the necessity of the locally path connected condition on $A$. The proof that $\tilde{A}$ is a covering space seems straightforward. We use the path-connectedness of $A$ obviously to talk about a single $\pi_1$. For the other part, take $i$ as the inclusion map of $A$ in $X$. If we have $[f]\in p_*(\pi_1(\tilde{A}))$, we have a homotopy to the constant map already since $\tilde{X}$ is simply connected, so $[f] \in \ker(i_*)$. For any $[f] \in \ker(i_*)$, by Proposition 1.31 the image of $f$ in $X$ lifts to some loop $\tilde{f}$ in $\tilde{X}$ with basepoint in $\tilde{A}$. Here the path connectedness of $\tilde{A}$ is useful. Then $[p(\tilde{f})] = [f]$, so $[f] \in p_*(\pi_1(\tilde{A}))$. And that should be it?