Proposition 4.16.2. Let $X$ be a topological space and $E, F \subset X$ be compact, then the following sets are compact:
$E \cup F$.
$G \subset E$ closed.
For any topological space $Y$ and $f \in C(E; Y)$, $f(E)$.
Proof. (1): Let $\seqi{U}$ be an open cover of $E \cup F$, then there exists $J_{1}, J_{2} \subset I$ finite such that $\bigcup_{j \in J_1}U_{j} \supset E$ and $\bigcup_{j \in J_2}U_{j} \supset F$. In which case, $\bigcup_{j \in J_1 \cup J_2}U_{j} \supset E \cup F$.
(2): Let $\seqi{U}$ be an open cover of $G$. Since $G$ is closed, $\seqi{U}\cup \bracs{G^c}$ is an open cover of $E$. In which case, there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_{j} \cup G^{c} \supset E$, so $\bigcup_{j \in J}U_{j} \supset G$.
(3): Let $\seqi{U}$ be an open cover of $f(E)$, then $\bracs{f^{-1}(U_i)}_{i \in I}$ is an open cover for $E$. Thus there exists $J \subset I$ finite such that $\bigcup_{j \in J}f^{-1}(U_{j}) \supset E$, so $\bigcup_{j \in J}U_{j} \supset f(E)$.$\square$