Say $X$ is a compact metric space, with $f:X \to X$ continuous. Now, for any $x_0$, $f^n(x_0)$ must have a convergent subsequence, say $f^{n_i}(x_0) \to x_\infty$. If we know that any such limit is a fixed point of $f$, that is, $f(x_\infty) = x_\infty$ for any such convergent subsequence, does it force the whole sequence $f^n(x_0)$ to converge to $x_\infty$?
If needed, we may assume $f$ has only finitely many, and hence discrete fixed points.
I see that each translate of the subsequence ($f^{n_i+r}(x_0)$) converges to $x_\infty$, and this would have solved the problem if the differences between consecutive $n_i$ were bounded. Also, I'm pretty sure that this, combined with the fixed points being isolated, forces any convergent subsequence to have the same limit, but that doesn't seem to prove the result either.
Edited: I realized my wording was wrong on a crucial point, and since the sole existing answer was not perfectly correct, I decided to put in the change. Previously we only knew the limit to be a fixed point for one particular subsequence, now we now this for any convergent subsequence of iterates.
Best Answer
Let $X$ be the subset of the Euclidean plane $\mathbb R^2$ consisting of the following points. For each $n=1,2,\dots$, let $X$ contain the points $(\frac kn,\frac 1n)$ for $k=0,1,\dots,n$; in addition. let $X$ contain all the points $(x,0)$ for $0\leq x\leq 1$. Define $f:X\to X$ as follows. Each of the points $(x,0)$ is fixed by $f$. For the remaining points, $f(\frac kn,\frac 1n)=(\frac{k+1}n,\frac1n)$ as long as $k<n$ while $f(1,\frac1n)=(0,\frac1{n+1})$. Then for any $p\in X$ not of the form $(x,0)$, the sequence $f^n(p)$ has subsequences converging to all the fixed-points $(x,0)$.
EDIT: Thanks to dfeuer for pointing out the error in my argument and for suggesting a correction. My construction was defined with $X$ included in the unit square $[0,1]^2$, and the error is that points of the form $(1,\frac1n)$ at the right edge are mapped by $f$ all the way over to the left edge, while their limit, $(1,0)$ is fixed by $f$. The problem goes away if we (1) identify the left and right edges of the square to form a cylinder and (2) delete the now redundant points on the right edge. To be specific, we'll have $f$ mapping each of the $n$ points "at level $n$", namely $(\frac kn,\frac 1n)$ for $k=0,\dots,n-1$, to the next one except that the last one $(\frac{n-1}n,\frac1n)$ gets mapped to $(0,\frac1{n+1})$, which is now, thanks to the wrap-around in the cylinder, a very small distance away.