From Intro to Topological Manifolds by Lee. I am confused about the proof of the closed subset part, which starts at line $4$ of the proof.
If $\overline{U \cap Z}$ is compact in $\bar U$, how does it follow that $\overline{U \cap Z}$ is compact in $Z$?
On
If $X$ is a topological space, if $Y\subset X$, if you consider in $Y$ the subspace topology, and if $K\subset Y$, then $K$ is compact in $Y$ if and only if $K$ is compact in $X$. So, yes, if $\overline{U\cap Z}$ is compact in $U$, then it is also compact in $\overline Z=Z$.
On
Suppose $\overline {U \cap Z} $ is covered by $\{V_i \cap Z\}$, $i \in I$ where each $V_i$ is open in $X$. Intersecting with $\overline {U}$ we see that $\overline {U \cap Z} \subset \cup_i \{V_i \cap Z \cap \overline {U}\} \subset \cup_i \{V_i \cap \overline {U}\} $. By hypothesis there is a finite subcover (since the sets in the union are open sets in $\overline {U}$). This gives the desired finite subcover of the original cover since $V_i \cap \overline {U} \subset V_i $.
Compactness is an absolute, not relative property, in the following sense:
(1). A space $W$ is compact iff every cover of $W$ by open subsets of $W$ has a finite sub-cover.
(2). A sub-space $W$ of a space $Y$ is compact in $Y$ iff every cover of $W$ by open sets of $Y$ has a finite sub-cover.
(3). The important point is that a space $W$ is compact iff $W$ is compact in any space $Y$ of which $W$ is a sub-space.
So if $\overline {U\cap Z}$ is compact in $\overline U$ then the $space $ $\overline {U\cap Z}$ is compact, and it is a sub-space of $\overline Z=Z,$ so it is compact in $Z.$
This implicitly uses (twice) the transitivity of the relation "sub-space": If $B$ is a sub-space of $X$ and if $C$ is a sub$set$ of $B$ then the topology on $C$ as a sub-space of $B$ is equal to the topology on $C$ as a sub-space of $X$.... (With $C=\overline {U\cap Z}$ and with $B=\overline U$ or with $B=\overline Z$.)