About a Path-connected Covering Space of a Subspace

Definition (Covering Space).
Let $X$ be a topological space. Then an arrangement $p:\cE\to X$ is a covering space of $X$ iff each point $x\in X$ has an evenly covered neighborhood $U$, i.e. an open neighborhood of $x$ such that
$$p^{-1}(U)=\bigsqcup_{i\in I} \cV_i$$
where $\{\cV_i\}_{i\in I}$ is a family of open sets in $\cE$, each of them mapped homeomorphically by $p$ into $U.$

A simply connected covering space is called an universal cover.
Proposition Nyaa.
Let \(p:\cE\to X\) be the universal cover of a (path-connected, locally path-connected and semilocally simply connected) space \(X\), and \(A\subseteq X\) path-connected and locally path-connected. Let \(\cA\subset\cE\) be a path-connected component of  \(p^{-1}(A)\). Then \(p:\cA\to A\) is the covering space corresponding to the subgroup \(\ker i^*\), where \(i^*:\pi_1(A,a)\to \pi_1(X,a)\) is the morphism induced by the inclusion $i:A\to X$.

Remark. Note that \(\cA\) is the covering space corresponding to a subgroup \(H\leq G\) when it can be said that $[\alpha]\in H$ iff \(\alpha\) lifts to a loop in \(\cA\). Note also that in the definition of a covering space (Taken from Hatcher's book, for example) it isn't said that a covering space is surjective.

Lemma 1. Let $p:\cE\to X$ be a covering space. Then $p$ is an open map.

Proof of Lemma 1.
Let $\cU$ be an open set in $\cE$. Let $x\in p(\cU)$. Let $U$ be an evenly covered neighborhood of $x$, and $\cV$ an open set of $\cE$ mapped homeomorphically by $p$ into $U$, such that $\cU\cap \cV\neq \varnothing$. Then since $p|_{\cV}$ is an homeomorphism, and $\cU\cap \cV$ is open in $\cV$, we get $p(\cU\cap\cV)$ is open in $U$, and also contained in $p(\cU)$. But since $U$ is open in $X$, $p(\cU\cap\cV)$ is also open in $X$: It's a neighborhood of $x$ contained in $p(\cU)$. Therefore, $p(\cU)$ is open, and we're done.

Lemma 2.
Let $p:\cE\to X$ a covering space and $A\subseteq X$ path-connected and locally path-connected. Let $\cU$ be a path-connected component of $p^{-1}(A)$. Then the restriction $\overline{p}:\cU\to A$ is a covering space, which also is surjective.

Proof of  Lemma 2.
It's easy to see that $\overline{p}:\cU\to A$ is a covering space. Indeed, take an evenly covered open neighborhood $W$ of a point $x\in A$ and intersect it with $A$. Then $\p^{-1}(W)=p^{-1}(W)\cap \cU$ is a (possibly empty) disjoint union of open sets in $\cU$ mapped homeomorphically by $p$ to $W$. Note that the same reasoning shows that $p:p^{-1}(A)\to A$ is a covering space.

Now let's see that $\p$ is surjective. Since a covering map is open, $\p(\cU)$ is an open set. Let's now suppose that $x\in A-\p(\cU)$. Let $U$ be a path-connected evenly covered neighborhood of $x$. Then $p^{-1}(U)$ is a disjount union of a family $\{\cV_i\}_{i\in I}$ of path-connected open sets of $\cE$, each of them mapped homeomorphically into $U$, each of them contained in a path-connected component of $p^{-1}(A)$, and each of them contained in $p^{-1}(A)-\cU$. Then $\p(\cU)\cap A=\varnothing$. So $A-\p(\cU)$ is open. Since $A$ is path-connected, in particular it's connected and $\p(\cU)=A$.

Proof of Proposition Nyaa.
We have that $[\alpha]\in\ker i^*$ iff $i^*([\alpha])=[\alpha]_X=0$ iff $\alpha$ is nullhomotopic in $X$ iff the lifting $\tilde{\alpha}$, starting at any point of $p^{-1}(a)$ in $\cA$, is nullhomotopic. But to be nullhomotopic, it must be a loop (More formally, the homotopy between $\alpha$ and a constant loop lifts to a homotopy between $\tilde{\alpha}$ and the constant loop starting at the same point, which makes $\tilde{\alpha}$ a loop), and since $\cE$ is simply connected, $\tilde{\alpha}$ is nullhomotopic iff it's a loop, which happens iff $[\alpha]\in \pi_1(E,\alpha(0))$. The result follows.