Let $X$ be a metric space and let $E\subset X$
Consider the property $\forall x\in X,\forall \delta>0 \ E\cap B(x,\delta)\ne\emptyset$
Can $E$ have zero measure?
Yes: there is $X=\Bbb R$ and $E=\Bbb Q$ where $E$ has both properties.
Can $E$ be not dense in $X$? If that's the case than there is a point $y$ in $X$ and no sequence in $E$ that converges to that point (by looking at the closure of $E$ as all the limit points of sequences of $E$). But then there is also a neighborhood of that point without any sequence in $E$ converging to any point in that neighborhood. Because again by contradiction if for any neighborhood of $V$ of $y$ there is a sequence in $E$ that converges to some point in $V\backslash\{y\}$ than we can find a sequence of sequences by taking smaller and smaller neighborhoods and from these we can build a sequence that converges to $y$ itself.
So is the property equivalent to $E$ being dense?
Yes, it is. Let $y\in E$ take $\delta_n={1\over n}, x_n\in E, x_n\in B(y,{1\over n})$, $x_n$ converges towards $y$.