Proof of Continuous Image of a Compact Space is Compact

Let $(V_i)_{i\in I}$ be an open cover of $f(X)$ in $Y$, so that $I$ is a set, $V_i\in\mathcal{T}_Y$ for every $i\in I$, and $f(X)\subseteq\bigcup_{i\in I}V_i$.

For $i\in I$ put

$$W_i=f^{-1}(V_i)=\{x\in X : f(x)\in V_i\}.$$

Since $f$ is continuous and $V_i\in\mathcal{T}_Y$, we have $W_i\in\mathcal{T}_X$ for every $i\in I$, so $(W_i)_{i\in I}$ is a family of subsets of $X$ consisting of open sets.

This family covers $X$. Indeed, let $x\in X$. Then $f(x)\in f(X)$, hence $f(x)\in V_i$ for some $i\in I$, and therefore $x\in W_i$. Thus

$$X\subseteq\bigcup_{i\in I}W_i.$$

By hypothesis $X$ is compact, so there is a finite subset $J\subseteq I$ with

$$X\subseteq\bigcup_{j\in J}W_j.$$

We claim that $f(X)\subseteq\bigcup_{j\in J}V_j$. Let $y\in f(X)$, and pick $x\in X$ with $y=f(x)$. Then $x\in W_j$ for some $j\in J$, that is, $f(x)\in V_j$, so $y\in V_j$. This proves the claim, so $(V_j)_{j\in J}$ is a subcover of $f(X)$ indexed by a finite subset of $I$.

Since $(V_i)_{i\in I}$ was an arbitrary open cover of $f(X)$ in $Y$, statement 2 of Compact Subset Criterion via Open Covers in the Ambient Space holds for the subset $f(X)\subseteq Y$. By that theorem, $f(X)$ is compact in $Y$.