Proof of Continuous Image of a Compact Space is Compact

Assume that $X$ is compact, and let $f:X\to Y$ be continuous. By Compact Subset Criterion via Open Covers in the Ambient Space, it is enough to show that every open cover of $f(X)$ in $Y$ has a finite subcover.

Let $I$ be a set, and let $(V_i)_{i\in I}$ be an open cover of $f(X)$ in $Y$. By Open Cover and Subcover of a Subset of a Topological Space, each $V_i$ is open in $Y$ and

For each $i\in I$, define

Since $f$ is continuous, each $U_i$ is open in $X$ by Continuous Map Between Topological Spaces. We claim that $(U_i)_{i\in I}$ is an open cover of $X$ in $X$. Let $x\in X$. Then $f(x)\in f(X)$, and because $(V_i)_{i\in I}$ covers $f(X)$, there exists some index $i\in I$ such that $f(x)\in V_i$. Hence $x\in U_i$. Therefore

So $(U_i)_{i\in I}$ is an open cover of $X$ in $X$.

Apply Compact Subset Criterion via Open Covers in the Ambient Space with the subset $X\subseteq X$. Since $X$ is compact, there exist a natural number $n\in\mathbb{N}$ and indices $i_1,\dots,i_n\in I$ such that

We show that $(V_{i_1},\dots,V_{i_n})$ covers $f(X)$. Let $y\in f(X)$. Then $y=f(x)$ for some $x\in X$. Since $X\subseteq U_{i_1}\cup\cdots\cup U_{i_n}$, there exists $k\in\{1,\dots,n\}$ such that $x\in U_{i_k}$. By the definition of $U_{i_k}$, this means that $f(x)=y\in V_{i_k}$. Hence

Therefore $(V_{i_1},\dots,V_{i_n})$ is a finite subcover of $f(X)$ in $Y$.

So every open cover of $f(X)$ in $Y$ has a finite subcover. By Compact Subset Criterion via Open Covers in the Ambient Space, the subset $f(X)$ is compact in $Y$.