Does $S^\bot+T^\bot = (S\cap T)^\bot$ hold in infinite-dimensional spaces?
It does not hold in general.
Recall two basic facts: for any subspace $E$ of a Hilbert space, we have that
$E^{\perp}$ is closed;
$E^{\perp\perp} = \bar{E}$, the closure of $E$. In particular, if $E$ is closed then $E^{\perp\perp}=E$.
Now in The direct sum of two closed subspace is closed? (Hilbert space), you can find an example of two closed subspaces $X_1, X_2$ of a Hilbert space $H$, such that $X_1 \cap X_2 = 0$ but $X_1 + X_2$ is not closed. Taking $S = X_1^\perp$, $T = X_2^\perp$, we have by fact 2 that $S^\perp = X_1$ and $T^\perp = X_2$. So $S^{\perp} + T^{\perp} = X_1 + X_2$ is not closed. On the other hand, by fact 1 $(S \cap T)^\perp$ is necessarily closed. In this case, since $S \cap T = 0$, we have $(S \cap T)^\perp = H$ which is definitely closed. So $S^\perp + T^\perp \ne (S \cap T)^\perp$.