Suppose that $(V,\|\cdot\|)$ is a Banach space. Let $S_n$ be a sequence of closed sets such that $\bigcup_{n\in\mathbb{N}} S_n = V$ and $S_n \subseteq S_m$ for $n\leq m$.
Let $(x_n)_{n\in\mathbb{N}}$ be a converging sequence in $V$ and define a new sequence $(\hat{x}_n)_{n\in\mathbb{N}}$ where each $\hat{x}_n\in S_n$ such that $\|x_n - \hat{x}_n\| = \min_{x\in S_n}\|x_n-x\|$.
Does it follow that $\| x_n - \hat{x}_n\|\rightarrow 0$?
I suppose you meant that $\hat {x_n} \in S_n$ and $\|x_n-\hat {x_n}\|=\inf_{y \in S_n} \|x_n-y\|$. Let $x_n \to x$. Then $x \in S_k$ for some $k$. For $n \geq k$ we have $x \in S_n$ so $\|x_n-\hat {x_n}\|\leq \|x_n-x\| \to 0$.