Let $X$ be a compact metric space and $Y$ any metric space. If $f:X \to Y$ is continuous, then $f(X)$ is compact (that is, continuous functions carry compact sets into compact sets).
Proof:
Consider an open cover of $f(X)$.
Then $f(X) \subset \bigcup_{\alpha \in A}V_\alpha$ where each $V_\alpha$ is open in $Y$.
$X \subset f^{-1}(f(X)) \subset f^{-1}\left(\bigcup_{\alpha \in A}V_\alpha\right) = \bigcup_{\alpha \in A}f^{-1}(V_\alpha)$.
Hence $\bigcup_{\alpha \in A}f^{-1}(V_\alpha)$ is an open cover of $X$. Since $X$ is compact then we can choose a finite subcover $\{V_i\}_{i=1}^n$ such that $X \subset \bigcup_{i=1}^n f^{-1}(V_i)$.
So then $f(X) \subset f\left(\bigcup_{i=1}^n f^{-1}(V_i)\right) = \bigcup_{i=1}^n f\left(f^{-1}(V_i)\right) \subset \bigcup_{i=1}^n V_i$, a finite subcover of $f(X)$. $\therefore f(X)$ is compact.
Does this proof have an error?
Thanks for your help.