Theorem [1, 2] Suppose $f\colon X\to Y$ is a continuous map between topological spaces $X$ and $Y$. If $X$ is compact and $f$ is surjective, then $Y$ is compact.

The inclusion map $[0,1]\hookrightarrow[0,2)$ shows that the requirement for $f$ to be surjective cannot be omitted. If $X$ is compact and $f$ is continuous we can always conclude, however, that $f(X)$ is compact, since $f\colon X\to f(X)$ is continuous.

Proof of theorem. (Following [1].) Suppose $\{V_{\alpha}\mid\alpha\in I\}$ is an arbitrary open cover for $f(X)$. Since $f$ is continuous, it follows that

is a collection of open sets in $X$. Since $A\subseteq f^{{-1}}f(A)$ for any $A\subseteq X$, and since the inverse commutes with unions (see this page), we have

Thus $\{f^{{-1}}(V_{\alpha})\mid\alpha\in I\}$ is an open cover for $X$. Since $X$ is compact, there exists a finite subset $J\subseteq I$ such that $\{f^{{-1}}(V_{\alpha})\mid\alpha\in J\}$ is a finite open cover for $X$. Since $f$ is a surjection, we have $ff^{{-1}}(A)=A$ for any $A\subseteq Y$ (see this page). Thus

Thus $\{V_{\alpha}\mid\alpha\in J\}$ is an open cover for $f(X)$, and $f(X)$ is compact. $\Box$

A shorter proof can be given using the characterization of compactness by the finite intersection property:

Shorter proof. Suppose $\{A_{i}\mid i\in I\}$ is a collection of closed subsets of $Y$ with the finite intersection property. Then $\{f^{{-1}}(A_{i})\mid i\in I\}$ is a collection of closed subsets of $X$ with the finite intersection property, because if $F\subseteq I$ is finite then

which is nonempty as $f$ is a surjection. As $X$ is compact, we have

and so $\bigcap_{{i\in I}}A_{i}\neq\varnothing$. Therefore $Y$ is compact. $\Box$