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> Let $S\neq \emptyset \subseteq \mathbb{R}^n $, prove or disprove: $x \in \overline{S} \iff B\_{\delta}(x) \cap S \neq \emptyset $ for any $\delta > 0$
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Here $\overline{S}$ denotes the closure of set $S$, $\overline{S} = S \cup S\_{L} $ . $S\_{L}$ denotes the set of all limit/accumulation points of $S$. 

**My attempt** : $(\Rightarrow)$ Let $x \in \overline{S}$. Which means $x \in S $ or $ x \in S\_{L} $, which further implies that $\exists \, B\_{\delta}(x)$ such that $ B\_{\delta}(x) \cap S \neq \emptyset $ for any $\delta > 0$. $ (\Leftarrow )$ Now let $B\_{\delta}(x) \cap S \neq \emptyset $ for any $\delta > 0$. This implies $x \in S$ or $x \in S\_{L} \implies x \in \overline{S} $. 

I am not sure why $B\_{\delta}(x) \cap S \neq \emptyset $ implies that $x$ should be in $S$ or $\overline{S}$. Is this proof correct ? Are there any other ways to solve this ?