If $U\subset W\subset V$ and $W\cap U^\perp = \{0\}$ then $U=W$

Question

Statement :

Let $V$ be a Hilbert space.
Let $U\subset W\subset V$ be closed subspaces.
Suppose that $W\cap U^\perp = \{0\}$.
Then $U=W$.

I know this is true in the finite dimensional case (see proof below).
Is this true in the infinite dimensional case?
Under which added hypothesis is it true?
Or is there a counter-example?

Proof :

Because $W$ is closed in the Hilbert space $V$, we have $V=W\oplus W^\perp$.
To show that $W=U$, it suffices to show that $V=U\oplus W^\perp$.
That is, we have to show two things :

1. $U\cap W^\perp = \{0\}$
2. $U + W^\perp = V$

The first equality is direct :
$$
U\subset W \implies U\cap W^\perp \subset W\cap W^\perp = \{0\}
$$
It remains to show the second equality.
We have :
$$
W\cap U^\perp = \{0\} \\
\implies (W\cap U^\perp)^\perp = \{0\}^\perp \\
\implies W^\perp + U^{\perp\perp} = V \quad \text{(if $V$ finite dimensional)} \\
\implies W^\perp + U = V \quad \text{(because $U$ closed)} \\
\implies U + W^\perp = V
$$
QED

Answer 1

Here is a proof. Let $w \in W$ be given. Let $u$ be the projection of $w$ onto $U$.

Then, $w - u \in W$ since $u,w \in W$ and $w - u \in U^\perp$ by the properties of the projection. Hence, $w - u = 0$ and this shows $w \in U$.