= Solution
A <covering map> is a continuous surjection $p:\widetilde X\to X$ for which each point has an open neighbourhood $U$ with $p^{-1}(U)$ a disjoint union of open sets, each mapped homeomorphically onto $U$. A connected simply connected <covering space> is a <universal cover>.
There is a necessary connectedness qualification in the printed conclusion. Let $A_0$ be the path component of $a_0$ in $A$. Then $p(\widetilde A)=A_0$: one inclusion follows by projecting paths, and the other by lifting a path in $A_0$ from $a_0$. Thus the asserted surjection onto all of $A$ requires $A$ to be path connected. Without it, take $p$ to be the identity of $\mathbb R$ and $A$ to be two disjoint intervals; a selected component cannot cover both.
To prove the assertion about a <component of a pulled-back universal cover>, choose an evenly covered neighbourhood $U$ in $X$. Local path connectedness of $A$ supplies a path-connected relatively open neighbourhood $V\subseteq U\cap A_0$. Each inverse sheet over $V$ is path connected, so it lies wholly in $\widetilde A$ or is disjoint from it. These sheets evenly cover $V$ under $p|_{\widetilde A}$. Therefore \b[$p|_{\widetilde A}:\widetilde A\to A_0$ is a <covering map>].
Under the <classification of connected covering spaces>, its <subgroup> consists of based loops in $A_0$ whose lifts close at $\widetilde a_0$. A loop viewed in $X$ closes in the <universal cover> exactly when its class in $\pi_1(X,a_0)$ is trivial. Hence the <subgroup> is
$$
\boxed{p_*\pi_1(\widetilde A,\widetilde a_0)=\ker\{\pi_1(A,a_0)\xrightarrow{i_*}\pi_1(X,a_0)\}.}
$$
Here $\pi_1(A,a_0)=\pi_1(A_0,a_0)$, so the <subgroup> formula itself needs no change. Injectivity of the covering's induced fundamental-group map is the usual homotopy-lifting property.
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