Is the proof for the following a valid proof? This proof has a different approach from others.
A metric space in which every infinite set has a limit point is separable.
Proof Let $E$ be a countably infinite subset of $X$ and $y$ be a limit point of $E$. $N_{\epsilon}(y)$ contains a point $p \in E$ $\forall \hspace{2pt} \epsilon > 0$. For any $x \in X$, let $d(x,y)=\delta$. Then $p \in N_{\delta+\epsilon}(x)$ which implies $x$ is also a limit point. So every point of $X$ is either a limit point of E or a point of E or both and hence E is a countable dense subset making X separable.
Best Answer
Just looking at the first and last sentences of your proof, it can't possibly be correct:
If this is correct, you've proven that every countable subset of a topological space with your property is dense. But notice that the set $\mathbb R\cup\{\infty,-\infty\}$ has your property, but not every countable subset of it is dense.
The mistake is when you conclude that $x$ is a limit point. Now, it's not obvious to me why you say this (you just kind of say it with no proof), but it isn't true. For example, $y=1$ is a limit point of $E=(0, 1)$, but $x=1,000,000,000,000$ is not a limit point of $E$.