Clousure of sum of two closed Hilbert subspaces with equivalent basis Let $H$ be an Hilbert space over $\mathbb{C}$
Let $V$ and $U$ be two closed subspaces of $H$
Let $\{v_m\}_{m \in \mathbb{N}} \in V$ be a basis for $V$
Let $\{u_m\}_{m \in \mathbb{N}} \in U$ be a basis for $U$
We know that $\{v_m\}_{m \in \mathbb{N}}$ is equivalent to  $\{u_m\}_{m \in \mathbb{N}}$ that is: there exists a topological isomorphism $T : V \to U$ such that $\forall m \in \mathbb{N}: T(v_m) = u_m$
My question is if is it true that: $V+U$ is closed
Thanks

I suppose you are talking about orthonormal bases. No infinite dimensional Hilbert space has  a countable Hamel basis. 
Any two separable infinite dimensional Hilbert spaces are isometrically isomorphic so your assumption of existence of a topological isomorphism is very weak. Any well known example of two infinite dimensional separable Hilbert subspaces whose sum is not closed would be  a counterexample. 
See Direct sum of two closed subspaces of Banach space is not closed
