Proof verification: Property of specific continuous map on compact metric space

Question

Assume $X$ is a compact metric space with metric $d$ and that $f:X\to X$ is a continuous map, such that $f(x)\neq x$ for any $x\in X$.
I want to prove $\exists\epsilon>0$ such that $d(f(x),x)\geq\epsilon$ for all $x\in X$.
My intial thought was to simply do this through a property of the metric $d$. Namely, $d(x,y)=0 \iff x=y$. We have by assumption that $f(x)\neq x$. Hence it follows that $d(f(x),x)\neq0$. But since $d:X\times X\to [0,\infty)$, it must be the case that $d(f(x),x)>0$, hence proving the existence of $\epsilon$.
However, I have not used the fact that $f$ is a continuous map at all. Have I made a mistake somewhere? I don't see a flaw in my reasoning, so if anyone could point out where I went wrong, or if the continuity of the function is just a red herring, I would very much appreciate that.

Answer 1

Hint

The claim is false. Consider $X=\mathbb{R}$ with $d(x,y)=|x-y|$ and $f(x)=x+e^x.$

Note that $d(x,f(x))>0,\forall x\in X,$ doesn't imply $\inf_{x\in X}\{d(x,f(x)\}>0.$

Added

In case $X$ is compact we have that $x\in X\to d(x,f(x))\in [0,\infty)$ is a continuous function. So, it achieves a minimum. Since $d(x,f(x))>0,\forall x\in X,$ the minimum value must be positive.