$X$ a metric space, proof that every infinite subset of $X$ has a limit point in $X$ given subsequence convergence Let $X$ be a metric space.

Assertion: Every sequence in $X$ has a subsequence that converges to a point of $X$.

Give the above Assertion, proof the following statement:

Every infinite subset of $X$ has a limit point in $X$.

Proof: Let $A\subset X$ be an infinite subset. Choose a sequence $(x_n)_{n=1}^\infty$ such that $x_n\in X$ for all $n$. Then by the Assertion, $(x_n)$ has a subsequence with a limit $x$ in $X$. This implies that $A$ has a limit point and thus the statement is proven.
Questions: Is this proof correct? I know that this is part of a Theorem about compact spaces and I need to prove this specific implication.
It is said that this implication can be proven such that it is valid in any topological space. Does the proof presented fulfills this? How can this be possible since $X$ is a metric space and thus it is implied that the topology of $X$ is the one induced by the metric?
 A: Your proof is not sufficient. You need to prove that any neighborhood of the limit $x$ contains at least one point of $X$ distinct of $x$.
And for that the hypothesis that $X$ is infinite is crucial.
A: Choosing a sequence in $X$ clearly won’t help you show that the set $A$ has a limit point: for that you need to choose a sequence in $A$. And you can’t choose just any old sequence: if $X=\Bbb R$, and $A=\{0\}\cup[1,2]$, the constant sequence $\langle 0,0,0,\ldots\rangle$ is a sequence in $A$ that converges to $0$, but $0$ is not a limit point of $A$, because $(-1,1)$ is an open nbhd of $0$ that contains no point of $A$ except $0$ itself.
You need to use the hypothesis that $A$ is infinite and start with a sequence $\langle x_n:n\in\Bbb N\rangle$ such that each $x_n\in A$, and $x_n\ne x_m$ whenever $n\ne m$, so that the points of the sequences are all distinct. Now use the hypothesis that this sequence has a subsequence converging to some $p\in X$ to show that $p$ must be a limit point of $A$.
