Continuous image of a compact set is compact

Statement

Let $X$ and $Y$ be topological spaces, let $f \colon X \to Y$ be a continuous map and let $K \subseteq X$ be compact. Then the image $f(K)$ is compact in $Y$.

Remarks

Proof: let $(V_i)$ be open sets of $Y$ that cover $f(K)$. The sets $f^{-1}(V_i)$ are open and cover $K$. A finite subfamily covers $K$, and the matching $V_i$ cover $f(K)$. A consequence is that a continuous real function on a compact set attains its bounds. Compactness is a topological property.

Explanations

L1 human

Continuous maps keep compactness. For example, the image of $[0, 1]$ under a continuous map into the plane is a compact curve, so it is closed and bounded.