Let's talk of inner product spaces.
I can show that if $V = U\oplus U^\perp$, then $(U^\perp)^\perp = U$ holds. Then I wonder about the converse.
I was able to dig up a post on SE where an example was given, showing that $U$ being closed is not enough for $V = U\oplus U^\perp$. I know that if $V$ is Hilbert, then $U$ being closed implies $(U^\perp)^\perp = U$. But the example's $V$ is not, rendering my search still open since there $U^\perp = \{0\}$ and $U\subsetneq V$.
But I am not able to find one, or disprove this. Hence I seek help! I'd appreciate if what you give is accessible with my current background.
My background: I am taking a course on linear algebra, and the professor deals only with finite dimensional spaces. But whenever a chance comes up, I try to generalise the results and see what fails in infinite dimensions. So far, I have seen some counterexamples via $\ell^2$.
Consider $V\subseteq\ell^2(\mathbb N)$ linearly spanned by all vectors with finite support and $w:=(1, \frac{1}{2}, \frac{1}{3}, \cdots)\in\ell^2(\mathbb N)$.
Let $U:=\{(a_i)\in V : \forall j\in \mathbb N, a_{2j}=0\}$.
We show that $U^{\perp}=\{(a_i)\in V: \forall j\in \mathbb N, a_{2j+1}=0\}$. For any $v=u+kw\in V$ where $u$ has finite support and $k$ is a scalar, we have if $k\not=0$, then for sufficiently large $n\ge N$, $v_n=kw_n\not=0$, and there is odd $n$ such that $(v_n, \mathbb{1}_{\{n\}})=v_n\not=0$. Now it's easy to see that an element $v=u$ with finite support is in $U^{\perp}$ iff its support lies in odd numbers.
Similarly, it can be shown $(U^{\perp})^{\perp}=U$. But $V\not=U\oplus U^{\perp}$ as $w\not\in U\oplus U^{\perp}$ which consists of only vectors of finite support.