Is the image of a closed subspace under a bounded linear operator closed? This seems obvious, but I can't get the proof straight, and I made up the statement myself, so I'm not sure if it's true in the stated generality. Given a bounded linear operator $T$ in Hilbert space, and a closed subspace $A$, I assert that $T(A)$ is a closed subspace. (Note: my definition of closed is in terms of sequences - the set contains all its limits under countable sequences - not the usual topology definition.)
Proof strategy: Let $f:\Bbb N\to T(A)$ be given. Using countable choice, select a function $g:\Bbb N\to A$ such that $T\circ g=f$. Assuming $g$ is convergent to some $x$, we have $x\in A$, and since $T$ is continuous, $T\circ g\to T(x)\in T(A)$.
The problem is that the preimage function $g$ may not converge even if $f$ does, and there is no limit on $\frac{|g(x)-g(y)|}{|f(x)-f(y)|}$ even though $T$ is bounded, because the inverse may not even exist so that there are whole subspaces to select very distant $g(x),g(y)$ even if $f(x),f(y)$ are close or even identical.
Is my search hopeless - have I bitten off more than I can chew with this statement, or can I somehow restrict the base sets from which the $g(x)$ were drawn so that $g$ will converge? What is the "correct" statement along these lines?

The statement is false.
Consider for example
$$T : \ell^2(\Bbb{N}) \rightarrow \ell^2(\Bbb{N}), (x_n)_n \mapsto (\frac{1}{n} \cdot x_n)_n .$$
Then $\rm{im}(T)$ contains the dense subspace $\rm{span}(e_1, e_2, \dots)$, where $e_1, e_2, \dots$ are the standard coordinate vectors.
But you can easily check that $(1/n)_n \in \ell^2(\Bbb{N})\setminus \rm{im}(T)$.
