Let $X$ be a separable normed space and $K$ closed convex subset of $X$. Prove that for some $(x_n^{\star})\subset X^\star$ and $(\lambda_n)\subset \mathbb{R}$ we have $\displaystyle K=\bigcap_{n=1}^{\infty}L_n$, where $L_n=\{x\in X: x_n^\star(x)\leq \lambda_n\}.$
Attempt. Since X is separable and $X\setminus K$ is open, we have $X\setminus K=\bigcup_{n=1}^{\infty}B(x_n,\epsilon_n)$ for some $x_n\in X,~\epsilon_n>0.$ By the separation theorem, for all $n$ we have $$\sup_{x\in K}x_n^\star(x)\leq \inf_{x\in B(x_n,\epsilon_n)}x_n^\star(x)$$ for some $x_n^\star\in X^\star$ and set $\displaystyle \lambda_n=\inf_{x\in B(x_n,\epsilon_n)}x_n^\star(x)$. Then for $L_n=\{x\in X: x_n^\star(x)\leq \lambda_n\}$, if $x\in K$ then $\displaystyle x_n^\star(x)\leq \sup_{x\in K}x_n^\star(x)\leq\lambda_n$ and $x\in L_n$ for all $n$. If $x\notin K$, then $x\in B(x_n,\epsilon_n)$ for some $n$ and $x_n^\star(x)\geq \lambda_n$.
This is where I am stuck: we would like to have $x_n^\star(x)>\lambda_n$, so $x\notin L_n$ for this $n$. But i don't seem to able to prove this.
Thanks in advance for the help.
You need to choose $x_n^{\ast} \neq 0$. That's possible by Hahn-Banach. Then, since $x_n^{\ast}\colon X \to \mathbb{R}$ is surjective, it is an open map, whence $x_n^{\ast}\bigl(B(x_n,\epsilon_n)\bigr)$ is an open subset of $\mathbb{R}$. And consequently, $x_n^{\ast}(x) > \lambda_n$ for all $x \in B(x_n,\epsilon_n)$.