Q: is there a convex bounded subset A of H such that A is not norm closed and A∩L is norm closed for every finite dimensional subspace L of H "Given an infinite dimensional Hilbert space H. Show that there is a convex bounded subset A of H such that A is not norm closed and A∩L is norm closed for every finite dimensional subspace L of H."
This question was my exam question. I cannot give an example such a subset A.

A: Assume $H$ is separable and $(e_n)_{n\in \mathbb{N}}$ is an orthonormal Hilbert basis of $H$. Let $S= \operatorname{span}(e_n)$ be the linear span of the basis vectors. Then $S$ is a proper subspace of $H$, and $\bar{S}=H$. Now let $A=S\cap \overline{B(0,1)}$, where $B(0,1)$ is the unit ball of $H$. We need to show that $A$ is convex and not closed.
$A$ is not closed since $S$ is a proper subset of $H$ and thus $A$ is a proper subset of $\overline{B(0,1)}$, which is the closure of $A$ since $S$ is dense. In short $A$ is not closed because $A\neq \bar{A}$.
Let $x,y\in A$, and $0<p<1$. Now $px +(1-p)y\in S$ because $S$ is a vector space, and $||px+(1-p)y||\leq p||x||+(1-p)||y||\leq 1$, and thus $px+(1-p)y\in \overline{B(0,1)}$. This shows that $A$ is convex.
Finally if $L$ is a finite dimensional subspace, then we know that $L$ is closed since every finite dimensional subspace in Hilbert space is closed. Now $L\cap A=L\cap S\cap \overline{B(0,1)}$. Since intersection of two subspaces is a subspace, $L\cap S$ is a finite dimensional subspace and thus closed. This shows that $A\cap L$ is closed since it is an intersection of two closed sets.
