Continuous Image of Compact Space is Compact

Theorem
Let $T_1$ and $T_2$ be topological spaces.

Let $f: T_1 \to T_2$ be a surjective continuous mapping.

If $T_1$ is compact then so is $T_2$.

That is, compactness is a continuous invariant.

Proof
Suppose $\mathcal U$ is an open cover of $f \left({T_1}\right)$ by sets open in $T_2$.

Because $f$ is continuous, it follows that $f^{-1} \left({U}\right)$ is open in $T_1$ for all $U \in \mathcal U$.

The set $\left\{{f^{-1} \left({U}\right): U \in \mathcal U}\right\}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $f \left({x}\right)$ must be in some $U \in \mathcal U$.

Because $T_1$ is compact, it has a finite subcover $\left\{{f^{-1} \left({U_1}\right), f^{-1} \left({U_2}\right), \ldots, f^{-1} \left({U_r}\right)}\right\}$.

It follows that $\left\{{U_1, U_2, \ldots, U_r}\right\}$ is a finite subcover of $T_2 = f \left({T_1}\right)$.