Does continuity implies contunity in the reversed sense?

Question

Let $f$ be a bijective real valued continous function with continous inverse defined on a closed interval $C$. Let $a\in C$. Then $\forall \epsilon>0, \exists \delta>0$ such that $\forall x \in C $ with $|x-a|<\delta$ we have $|f(x) - f(a)| < \epsilon$. Does it follow from this that $\forall \epsilon>0, \exists \delta>0$ such that $\forall x \in C$ with $|f(x) - f(a)| < \delta$ we have $|x-a|<\epsilon$ ? I believe the answer is yes and this has something to do with the inverse being continous but I couldn't fill in the gaps. I use this proposition to solve a homework question about finding completion of a metric space. I would love to see the proof of this proposition.