I'm working in the setting of complex, separable, infinite-dimensional Hilbert spaces and I came out with the following statement which, intuitively, seems to be true to me, though I haven't found a proof or a counterexample.
Let $H$ be a Hilbert space as above, and let $A_n$ be a sequence of closed subspaces such that $A_n^{\perp}$ has dimension $m \geq 1$ fixed for all $n \geq 1$. Moreover, assume that:
$$ \bigcap_{n=1}^{N} A_n $$
is infinite dimensional for all $N$. Is it true that:
$$ \bigcap_{n=1}^{\infty} A_n \neq 0$$
I thought of constructing a nested subsequence $B_n$ of closed sets of $A_n$ such that the diameter of $B_n$ tends to $0$ and all $x \in B_n$ satisfies $\| x \| =1$, so that we could apply Cantor's intersection Theorem and conclude that $\exists x :$
$$ \bigcap_{n=1}^{\infty} A_n = \lbrace x \rbrace$$
Then, since $x$ cannot be $0$ by hypothesis, the infinite intersection of all $A_n$ must be at least one-dimensional.
However, I have not found any sequence $B_n$ as above. Maybe we might take the closure of a basis set for $A_1$ and construct a basis for each $A_n$ removing the "unnecessary" vectors, and take $B_n$ to be the closure of these sets?
Counterexample: take $H=\ell^2$, and $A_n=\{x=(x_1,x_2,\ldots)\in\ell^2: x_1=\ldots=x_n=0\}$ ,then each $A_n$ is closed, with $\dim(A_n^{\perp})=n\ge1$, and any finite intersection among them is an infinite dimensional.
If $x\in\bigcap_{n=1}^{\infty}A_n$, and $x_k\neq0$, then $x\notin A_k, A_{k+1}, \ldots$, yielding a contradiction, thus no coordinate in x can be non-zero, which means the intersection can only be $0$