\begin{proof} If $\{ V_{\alpha} \}$ is an open cover of $f(K)$, then $\{ f^{-1}(V_{\alpha}) \}$ is an open cover of $K$, hence $$K\subset f^{-1}(V_{\alpha_{1}})\cup\dots \cup f^{-1}(V_{\alpha_{n}})$$ for some $\alpha_{1},\dots,\alpha_{n}$ and therefore (since $f(f^{-1}(V_{\alpha_{i}}))\subset V_{\alpha_{i}}$) $$\begin{align*} f(K)&\subset f\left( \bigcup_{i=1}^{n}f^{-1}(V_{\alpha_{i}}) \right)\\ &= f\left( \bigcup_{i=1}^{n}\{x \in X:f(x)\in V_{\alpha_{i}} \} \right)\\ &= f\left( \left\{ x \in X:f(x)\in \bigcup_{i=1}^{n}V_{\alpha_{i}} \right\} \right)\\ &= \bigcup_{i=1}^{n}V_{\alpha_{i}}. \end{align*}$$ \end{proof}