Why does a convex set have the same interior points as its closure? Let $C$ be a convex subset of $\mathbb{R}^n$. I've been trying for hours to prove that $\dot{\overline{C}}=\dot{C}$. Somehow my intuition completely fails me. I found a proof in a textbook, but just got stuck on another statement the author considered obvious. Could someone please give a proof that uses little more than elementary linear algebra, topology, and the definition of a convex set?
Edit: 
The proof mentioned above is from Blackwell and Girshick:
Let $y\in\dot{\overline{C}}$ and $T$ be a ball around $y$ contained in $\overline{C}$. Then $C\cap T$ has an inner point, as otherwise $C\cap T$ would be contained in a hyperplane and $\overline{C\cap T}=\overline{T}$ would be contained in the same hyperplane. The problematic statement is "as otherwise $C\cap T$ would be contained in a hyperplane".
Another thing: I would be interested in a proof that doesn't use the theorem about separating a convex set from a point by a hyperplane, as I came across this problem in a proof of that very theorem (in the appendix of Stochastic Finance by Föllmer and Schied). To be more precise, it occurs in the case of the point in question being in the boundary of $C$, when it is tacitly assumed, that it is also in the boundary of $\overline{C}$. I know this isn't strictly necessary, as I could use another proof, e.g. the one referred to by Mike, but now I'm curious.  

Interior points of a convex set $K$ in $\mathbf{R}^n$ are the points in what could be called the "convex interior" of the set: they are inside (the part with positive barycentric coordinates of) at least one nondegenerate n-dimensional simplex with vertices in $K$. 
Given a point in the interior of the convex set $\overline{C}$, surround it by a nondegenerate $n$-dimensional simplex with vertices in $\overline{C}$.  Because $C$ is dense in its closure, we can perturb the simplex very slightly into one with vertices in $C$, and this simplex will continue to contain the given interior point.  (This is because, for example, the barycentric coordinates are continuous functions of the simplex vertices, as long as the simplex does not degenerate, so a small perturbation will keep the coordinates positive and the point strictly inside the simplex).
