Opening and closing convex sets

Either use the suggestion by Andrea, or you can prove directly this way: $K \subseteq \bar K$, so $\mathring K \subseteq (\bar K)^\circ$. On the other hand, if $x \in (\bar K)^\circ$ then there is a neighborhood $x \in U_x \subseteq \bar K$. Now take a small simplex in $U_x$ that contains $x$ in its interior. Up to perturbing slightliy its vertices, you can make them to belong to $K$ while $x$ still belongs to the interior of the simplex. Then by convexity the simplex is in $K$, so $x \in \mathring K$.

I realized after writing that my answer is essentially the same as https://math.stackexchange.com/a/7509.

Here is a more functional-analytic proof, which avoids the argument with simplexes: assume by contradiction that there exists $x\in(\overline{K})^\circ\setminus K^\circ$. By Hahn-Banach theorem (applied to $K^\circ$ and $\{x\}$) and the fact that $(\mathbb{R}^n)^*=\mathbb{R}^n$, you can find $v\neq 0$ such that $\langle v,K^\circ\rangle \le \langle v,x\rangle$. So $$ \langle v,K^\circ\rangle + \epsilon|v|^2\le\langle v,x+\epsilon v\rangle. $$ But $x+\epsilon v\in\overline{K}$ if $\epsilon>0$ is small. In particular you can find $x'\in K$ such that $|(x+\epsilon v)-x'|<\frac{\epsilon}{2}|v|$, which gives $$ \langle v,K^\circ\rangle + \frac{\epsilon}{2}|v|^2\le\langle v,x'\rangle. $$ But by the "segment lemma", for any fixed $z\in K^\circ$, $[z,x')\subseteq K^\circ$, so $\langle v,[z,x')\rangle\le\langle v,x'\rangle-\frac{\epsilon}{2}|v|^2$ and by continuity $\langle v,x'\rangle\le\langle v,x'\rangle-\frac{\epsilon}{2}|v|^2$, contradiction.

We assumed $K^\circ\neq 0$, but you can always find an affine subspace containing $K$ where this happens and you can repeat the proof here.