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.
Suppose the sequence does not converge to $l$. Then $\exists$ an $\epsilon > 0$ such that for every $n_{0} \in \mathbb{N}, \exists n > n_{0}$ such that $d(x_{n},l) \geq \epsilon$. Thus, we get a subsequence $x_{n_{k}}$ such that $d(x_{n_{k}},l) \geq \epsilon$ for every $k \in \mathbb{N}$. Now using the compactness of the space, this subsequence has a convergent subsequence, say $x_{n_{k_{m}}}$, which cannot converge to $l$, however. But this subsequence is also a subsequence of the original subsequence and hence should converge to $l$. Hence the contradiction.
Best Answer
Let the original sequence be $\sigma$, and let $K=\operatorname{cl}\{x_n:n\in\Bbb N\}$; by hypothesis $K$ is compact. If $\sigma$ does not converge to $x$, there are an $\epsilon>0$ and a subsequence $\sigma_1=\langle x_{n_k}:k\in\Bbb N\rangle$ such that $d(x_{n_k},x)\ge\epsilon$ for all $k\in\Bbb N$. Clearly $\sigma_1$ is a sequence in the compact set $K$, so it has a convergent subsequence $\sigma_2=\langle x_{n_{k_i}}:i\in\Bbb N\rangle$. But $\sigma_2$ is also a subsequence of $\sigma$, so by hypothesis it converges to $x$, contradicting the fact that $d(x_{n_{k_i}},x)\ge \epsilon$ for each $i\in\Bbb N$. Thus, $\sigma$ must converge to $x$.