TITLE: Why does a convex set have the same interior points as its closure?
QUESTION [26 upvotes]: 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.

REPLY [9 votes]: This is from a set of class notes.
You need a lemma (interior and closure operators are $\mathrm{int},\mathrm{cl}$, resp.):
Lemma. If $C$ a convex subset of a topological space, and if $x\in \mathrm{int} C$ and $y\in \mathrm{cl} C$, then $[x,y)\subset \mathrm{int} C$.
The half-open segment $[x,y)$ is a right-open convex combination. If $\mathrm{int}C$ is empty, then the above lemma is true. Try to prove why this is true when the interior is not empty.
Proposition. $\mathrm{int} C = \mathrm{int}\;\mathrm{cl}\;C$.
Let $y\in \mathrm{int\; cl\;} C$; position it in an open ball, noting that the in the interior of the closure of $C$ is open, and hence $\exists r>0$ such that $B_r(y)\subset \mathrm{cl\;} C$. Pick a $y'\in \mathrm{int}\;C$. Then there exists an $\epsilon>0$, such that $y'+(1+\epsilon)(y-y') = y+\epsilon(y-y')\in B_r(y)\subset \mathrm{cl}\;C$. At the same time, $y$ belongs to the segment $[y',y+\epsilon(y-y'))$, so that, by the previous lemma, $y$ belongs to $\mathrm{int\;}C$. The reverse inclusion is straightforward, since $C\subseteq \mathrm{cl\;}C$, and thus obtaining the result.