Continuous images of compact sets are compact

Step 1 — Pull back an arbitrary open cover of $f(K)$. Let $\{V_i\}_{i \in I}$ be any collection of open sets in $Y$ covering $f(K)$, i.e. $f(K) \subseteq \bigcup_i V_i$. Since $f$ is continuous, each $f^{-1}(V_i)$ is open in $X$. For every $x \in K$, $f(x) \in f(K)$ lies in some $V_i$, so $x \in f^{-1}(V_i)$; thus $\{f^{-1}(V_i)\}_{i \in I}$ is an open cover of $K$.

Step 2 — Use compactness of $K$ to extract a finite subcover. Since $K$ is compact, finitely many of these sets already cover $K$: there exist $i_1, \dots, i_n$ with $K \subseteq f^{-1}(V_{i_1}) \cup \dots \cup f^{-1}(V_{i_n})$.

Step 3 — Push the finite subcover forward. Applying $f$ to both sides, $f(K) \subseteq f\big(f^{-1}(V_{i_1}) \cup \dots \cup f^{-1}(V_{i_n})\big) \subseteq V_{i_1} \cup \dots \cup V_{i_n}$ (using $f(f^{-1}(V)) \subseteq V$ always). So $\{V_{i_1}, \dots, V_{i_n}\}$ is a finite subcover of $f(K)$ drawn from the original cover. Since $\{V_i\}$ was arbitrary, $f(K)$ is compact.

Step 4 — The corollary: compact subsets of Hausdorff spaces are closed. Let $K \subseteq Y$ be compact, $Y$ Hausdorff, and fix any $y \notin K$. For each $x \in K$, Hausdorff separation gives disjoint open sets $U_x \ni x$ and $W_x \ni y$. The sets $\{U_x\}_{x \in K}$ cover $K$, so finitely many $U_{x_1}, \dots, U_{x_n}$ already cover $K$ by compactness. Then $W = W_{x_1} \cap \dots \cap W_{x_n}$ is a finite intersection of open sets (hence open), contains $y$, and is disjoint from every $U_{x_j}$, hence disjoint from $K \subseteq \bigcup_j U_{x_j}$. So every point outside $K$ has an open neighborhood missing $K$: the complement of $K$ is open, i.e. $K$ is closed.