Let $X$ be a compact, convex subset of $\mathbb R^n$ and let $f:X \to X$ be a continuous function.
Is it possible for $f$ to have countably, but not finitely many fixed points?
Putting it differently, let $g(x):=f(x)-x$. If $g$ is continuous, can it "cross 0" countably many times?
There are such functions.
To construct one, you can start by letting $\alpha$ (resp. $\beta$) denote the greatest fixed point (second greatest fixed point) of $h(x) = x + x \sin \left( \frac{1}{x} \right)$ on the interval $[-1,1]$.
It's clear that $h$ is continuous and has countably many fixed points, but it's not a map $X \rightarrow X$. Fortunately, we can make some changes to $h$ on the interval $[-\alpha, -\beta]$ and obtain a function $f: [-\alpha, \alpha] \rightarrow [-\alpha, \alpha]$ that has the same property: countably many fixed points, with $X=[-\alpha,\alpha]$ a compact, convex subset of $\mathbb{R}$.
I include a plot of $h$ (shaded square) and one possible $f$ below. The claimed properties are clearly visible.