I'm reading a proof of the Fredholm alternative, and there is a claim that goes like this: Let $K:X\rightarrow X$ be a compact linear map. Define $T=I-K$, then $Y=\ker(T)$ is a finite dimensional subspace with $X=Y\oplus Z$, where $Z$ is a closed subspace of $X$. Define $S:Z \rightarrow X$ by $Sx=Tx$. Then $S$ is injective. So far so good.
Here comes the part I don't get: If $\operatorname{range}(S)$ is not closed, then we have a sequence $(z_n) \subset Z$ such that $\|z_n\|=1$ for all $n$ and $\|Sz_n\|\rightarrow 0$. How does one construct such a sequence? The author appears to think this is obvious, so I must be missing something here.
If $\operatorname{range}(S)$ is not closed, then we can find a sequence $(y_n)_n\subset \operatorname{range}(S) $ such that $y_n\to y$ and $y\notin \operatorname{range}(S)$. We may write $y_n$ as $x_n-Kx_n$, where $x_n\in Z$.
Claim. The sequence $(x_n)_{n\geqslant 1}$ is not bounded.
Indeed, if it was, then the sequence $(Kx_n)_{n\geqslant 1}$ would admit a convergent subsequence, say $(Kx_{n_i})_i$. Since $x_{n_i} =Kx_{n_i} +y_{ n_i} $, the sequence $(x_{n_i})_i$ is convergent, say to $x$. But in this case, $y=Sx$, which would be a contradiction.
Now we pick a sequence of integers $m_j\uparrow \infty$ such that $\lVert x_{m_j} \rVert\gt j$ for each $j$, and define $z_j:=\left(\lVert x_{m_j}\rVert\right)^{-1}x_{m_j}$. Then $$S(z_j)=\frac 1{\lVert x_{m_j}\rVert}S(x_{m_j})=\frac 1{\lVert x_{m_j}\rVert}y_{m_i}, $$
from which it follows that $\lVert S(z_j)\rVert\leqslant \sup_i\lVert y_{i}\rVert /j$.