Let $V$ be a Hilbert space and let $\{V_h\}_{h>0}$ be a family of increasing subspaces being dense in $V$, i.e. \begin{align*} \overline{\bigcup_{h>0} V_h} = V. \end{align*} Given a closed and convex subset $K \subseteq V$ containing $0$, is the family $\{K \cap V_h\}_{h>0}$ dense in $K$? In other words does \begin{align*} \overline{\bigcup_{h>0} K \cap V_h} = K. \end{align*} hold?
I feel like the answer should be positive. Suppose the interior of $K$ is not empty. Given $v \in \mathring{K}$ there exists a sequence $\{v_h\}_{h>0}$ with $v_h \in V_h$ for any $h>0$ and $v_h \to v$ in $V$. As there exists a neighborhood of $v$ fully contained in $K$, there exists an index from which on the sequence will be contained in this neighborhood and therefore especially in $K$. What about points on the boundary of $K$ though?
$K$ need not have any interior, so your argument is false. Example: Let $w \in V$ which is not in any $V_h$ (this must exist by the Baire Category Theorem), and $K$ the line segment from $0$ to $w$.