Q: Nowhere dense sets. 
Given $X$ a metric space, $A\subset X$ a nowhere dense set. Show that every open ball $B$ contains another open ball $B_1 \subset B$ such that $B_1 \cap A = \emptyset$.

EDIT: I modify my proof with the help of a great hint!
Given $B(x,\epsilon)$ lets suppose that for every open ball $B_1 \subset B(x,\epsilon)$, $B_1 \cap A \neq \emptyset$. 
Given $y \in B(x,\epsilon)$ we have that for every $n>N$ with $N$ sufficiently big, $B(y,\frac{1}{n})\cap A \neq \emptyset$ That is, for every $n>N$ there exists $y_{n} \in B(y,\frac{1}{n})\cap A \neq \emptyset$. If we consider the sequence $(y_{n})_{n>N}$ we have that for every $n$, $y_n \in A$ and $y_n \rightarrow y$. Therefore $y \in cl(A)$ which implies $B(x,\epsilon)\subset cl(A)$ with contradicts the fact that $A$ is nowhere dense. 

That step is not legitimate. Consider the set $A=\left\{\frac1n:n\in\Bbb Z^+\right\}$ in $\Bbb R$. For each $\epsilon>0$ we have $B(0,\epsilon)\cap A\ne\varnothing$, but we can’t take the limit as $\epsilon\to 0^+$ to conclude that $\{0\}\cap A\ne\varnothing$: this is clearly false.
Try this instead. Suppose that every open ball contained in $B(x,\epsilon)$ intersects $A$. Show that this implies that $B(x,\epsilon)\subseteq\operatorname{cl}A$, contradicting the hypothesis that $A$ is nowhere dense in $X$.
