Prove $f^{-1} : f(A) \to \mathbb{R}$ is continuous Prove $f^{-1} : f(A) \to \mathbb{R}$ is continuous 
Let $f : A \to \mathbb{R}$ be an injective and continuous function and $A \subset \mathbb{R}$ is compact
Since f is continuous and $A$ is compact then $f(A)$ is compact
and since $f$ is injective, $\forall y \in f(A), \exists! x \in A \ni f(x) = y$
I'm not sure if this needs a $\varepsilon$-$\delta$  proof or some other proof.  

You haven't show that $f^{-1}$ is continuous.  It suffices to show that if $B\subseteq A$ is closed, then $f(B)$ is closed.  Since $B$ is a closed subset of a compact set, it is compact.  Therefore $f(B)$ is compact and hence closed.
