Equivalence between closed range + finite dimensional kernel, and a statement about sequences.

Question

I'm reading some notes on Fredholm operators, where the following lemma is provided, with a proof:

Lemma: the following are equivalent, given $T: X \to Y$ a bounded linear map between Banach spaces.

1. $\ker T$ is finite dimensional and $\text{im}\, T$ is closed.
2. Every bounded sequence $(x_i)$ in $X$ with $Tx_i$ convergent has a convergent subsequence.

The part of the proof I'm interested in is $2 \Rightarrow 1$, which is as follows:

"Now suppose that 2 holds. Then a bounded sequence in the kernel has a convergent subsequence so the kernel is finite dimensional. That Ran($T$) is closed follows immediately from 2."

However I don't understand why the last sentence is valid. If I wish to prove that the image of an operator is closed, I would start like this:

Suppose $(x_n)$ is any sequence in $X$, and suppose $Tx_n \to y$ in $Y$. We wish to prove that there exists some $x \in X$ such that $y = Tx$.

Why does it suffice to consider only bounded sequences $(x_n)$ rather than arbitrary sequences?

Answer 1

I don't see why it should follow "immediately" from $2$ but I can at least provide a proof.

Suppose that $\ker T$ is finite dimensional and that $\operatorname{im} T$ is not closed. There is a closed subspace $M$ such that $X = \ker T \oplus M$, the restriction $T\vert_M$ is injective but not an isomorphism. Now take a sequence $(z_n)$ in $M$ such that $\|z_n\| = 1$ and $\lim_n \|Tz_n\| = 0$. Then $(z_n)$ cannot have a convergent subsequence, because if $z$ were a limit of such a subsequence, then it would follow that $z \in M \cap \ker T$ and $\|z \| = 1$, which is a contradiction.