Let $B$ be a Banach space and $\left\{ X_{n}\right\} $ closed subspaces of $B $ with $X_{n}\neq B$ for any $n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $. Prove that $\underset{n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion }{\cup }X_{n}\neq B$.
I have only been able to show that $ X_{n}$ has empty interior, but I have failed to show that $\underset{n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion }{\cup }X_{n}\neq B$ I appreciate the attention.
Suppose to the contrary that $\underset{n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion }{\cup }X_{n}= B$. Baire says now: there is some $n \in \mathbb N$ such that $X_n$ has a non-empty interior.
Let $x_0 \in X_n^{o}$. Then there is $r>0$ such that
$K:=\{x \in B: ||x-x_0||<r\} \subseteq X_n$
Now let $z \in B$ and $z \ne x_0$ and put $t:=\frac{r}{2 ||x_0-z||}$ and $y:=x_0+t(x_0-z)$. Then
$||x_0-y||=t||x_0-z|| = r/2<r$. Hence $y \in K$ and therefore $y \in X_n$.
Since $X_n$ is a subpace and $x_0 \in X_n$:
$z=x_0+(x_0-y)/t \in X_n$.
We then have $X_n=B$, a contradiction.