Convex dense subset of $\Bbb{R}^n$ is the entire space

This question (Why does a convex set have the same interior points as its closure?) shows that the interior of $X $ is the same as the interior of the closure of $X $ (which is all of $\Bbb{R}^n $). Since a set contains it's interior, we are done.

Using induction over the dimension $n$, it is easy to show that, given any set of $2^n$ points meeting every quadrant of $\Bbb R^n$, the zero vector is in their convex hull.
Let $Q_i$ denote the $i$-th quadrant for $i=1,\dots,2^n$. Assume $a$ is a point in $\Bbb R^n$. Since $a+Q_i$ is open, we can pick a point $x_i$ from $X\cap(a+Q_i)$. Now since $a$ is in the convex hull of these $x_i$ and $X$ is convex, $a$ must be in $X$.