Prove that if $(M,\ d)$ is compact, every infinite subset of $M$ has a limit point.
Definitions:
$P$ is compact iff every sequence in $P$ has a subsequence that converges to a point in $P$.
Let $A$ be any subset of $(M,\ d)$. $x\in M$ is a limit point of $A$ iff there exists a sequence $\{x_n\}$ of points with $x_n\in A\ \forall n$, $x_n\ne x$ and $x_n\to x$.
My attempt:
Let $M$ be compact. Let $A$ be an infinite subset of $M$.
Let $\{x_n\}\subset A$ be composed of different elements of $A$ in any order. We know we can find such a sequence because $A$ is infinite.
$\{x_n\}\subset M$
Since $M$ is compact, $\{x_n\}$ has a subsequence $\{x_{n_i}\}$ that converges to a point $x\in M$. We claim that $x$ is a limit point of $A$.
Form a new sequence from $\{x_{n_i}\}$ after removing all the terms are equal to $x$. Since the terms of $\{x_{n_i}\}$ are all different, there can be at most one such term. The terms of this new sequence lie in $A$, are never equal to $x$ and converge to $x$. Thus, by our definition, $x$ is a limit point of $A$.
So, $A$ has a limit point.
QED
Could someone please go through this proof and tell me if I've missed something?