Proof check: existence of a sequence converging to a limit point

[Ozma]

Problem: let $E \subseteq \mathbb{R}^n$ and let $\mathbf{y}$ be a limit point of $E$. Prove that there exists a sequence $\mathbf{x}^h \ne \mathbf{y}$ of elements of $E$ such that $\mathbf{x}^h \to \mathbf{y}$ when $h \to +\infty$.

In the following work, with $B_r(\mathbf{a})$ I mean the ball of radius $r > 0$ and center $\mathbf{a} \in \mathbb{R}^n$.

My work: by hypothesis $\mathbf{y}$ be a limit point of $E$, hence for each $\epsilon > 0$ we have $(B_\epsilon(\mathbf{y}) \setminus \{\mathbf{y}\}) \cap E \ne \varnothing$. Being the intersection non-empty, there exists $\mathbf{x}^\epsilon \in (B_\epsilon(\mathbf{y}) \setminus \{\mathbf{y}\}) \cap E$. By definition of intersection, this means that $\mathbf{x}^\epsilon \ne \mathbf{y}$ and $\mathbf{x}^\epsilon \in E$. Let $h \in \mathbb{N}$, choosing $\epsilon = 1/(h+1)$, from $\mathbf{x}^{1/(h+1)} \in B_{1/(h+1)}(\mathbf{y})$ it follows that $\mathbf{x}^{1/(h+1)} \to \mathbf{y}$ when $h \to +\infty$ because, from the definition of ball in $\mathbb{R}^n$, we have $\| \mathbf{x}^{1/(h+1)}-\mathbf{y} \| < \frac{1}{h+1} \to 0$ when $h \to +\infty$.

Is my proof correct?

 

[blamocur]

Looks good to me. I am assuming that "the ball" means an open ball.

 

[Ozma]

@blamocur: Thank you for answering. Yes, I meant the open ball, sorry for the inaccurate writing.