If $U$ is closed, is Span($U \cup \{z\}$) also closed?

Question

I am trying to go through a proof of the Hahn-Banach Theorem for a separable Banach space $X$. I was hoping the following was true since it would be helpful:

If $X$ is Banach and $U$ is a closed subspace of $X$, then for any $ z \in X$ the subspace Span($U \cup \{z\}$) is also closed in $X$.

I tried proving it but didn't really get very far. Can anyone tell me if this is true, if so any hints for proving it?

Answer 1

Yes, its true.

Hint:

Suppose that $z \not\in U$. Using Hahn-Banach, you get $\varphi \in X'$, such that $\varphi(u) = 0$ for all $u \in U$ and $\varphi(z) \ne 0$.

Then, take a convergent sequence in $\operatorname{span}(U \cup \{z\})$ and apply $\varphi$ to this sequence.