Consider a metric space $X$ and its open subset $A$. $\newcommand{\ball}[2]{B_{#1}(#2)} \newcommand{\eps}{\epsilon}$
Proposition: $\forall x \in X: x \in \partial A \Rightarrow x \in L(A)$ where $L(A)$ is the set of limit points of $A$
My proof: Let $x$ be a boundary point of $A$. Then, $x \in \overline{A}\setminus A^\circ$, i.e., $x \in \overline{A}$ and $x \not\in A^\circ = A$ because $A$ is open. For all $\eps > 0$, there is $y \in A$ such that $y \in \ball{\eps}{x}$. From the fact that $x \not\in A$, we have that for all $\eps >0$, there is $y \in A$ such that $y \in \ball{\eps}{x}$ different from $x$.
Please check if my proof is correct.
Best Answer
Your proof is correct, since your last line is exactly the definition of limit point.
Just as a curiosity, note that every limit point in a metric space (first countable would be enough) can be reached by a convergent sequence. This property seems trivial, but may fail in general topological spaces where the topology is not induced by a metric. Then, more sophisticated tools like nets or ultrafilters are required to characterize the closure of a set.