For storytelling purposes, let me recall the theorem and the different steps of the proof. The question is in boldface after 2. followed by an "almost answer" in comments...
Let $\big(E,\lVert\,\cdot\,\rVert\big)$ be a Banach space, $K\in \mathcal{K}(E)$ a compact operator. Then $\operatorname{Im}(I-K)=\operatorname{Ker}(I-K^*)^{\perp}$ closed and $$ \operatorname{codim}\operatorname{Im}(I-K) = \operatorname{dim}\operatorname{Ker}(I-K) = \operatorname{dim}\operatorname{Ker}(I-K^*) < +\infty$$
(One finds the statement and proof in "Functional Analysis, Sobolev Spaces and Partial Differential Equations" (UTX 2011), Haim Brezis, Thm 6.6 p.160; or in these notes by Bernard Maurey, Thm 8.3.8 p.113; or in the case of Hilbert spaces in "PDE", Lawrence C. Evans, Thm 5 p.725)
- One first proves that $\operatorname{dim}\operatorname{Ker}(I-K) < +\infty$ by saying that on that subspace $K\equiv I$ (identity) so it maps the unit ball to itself. By def. of $K$ compact operator, the ball has to be relatively compact so the space $\operatorname{Ker}(I-K)$ has to be finite dimensional.
- Then in order to prove $\operatorname{dim}\operatorname{Im}(I-K)$ closed: by 1., there exists a finite basis $(\mathbf{e}_1,\cdots,\mathbf{e}_k)$ a basis of $\operatorname{Ker}(I-K)$. Denote $(\boldsymbol{\epsilon}^1,\cdots,\boldsymbol{\epsilon}^k)$ its dual basis. Extend each linear form from $\operatorname{Ker}(I-K)$ to $E$ (by Hahn-Banach, so that they are still continuous). This gives a supplementary to the kernel: $$\displaystyle E=\operatorname{Ker}(I-K) \oplus \underbrace{\bigcap_{i=1}^k \operatorname{Ker}\boldsymbol{\epsilon}^i}_{F}$$ $(I-K)|_F$ is injective and yields an isomorphism $F\cong \operatorname{Im}(I-K)$.
Question: Doesn't this already show $E/\operatorname{Im}(I-K) \cong \operatorname{Ker}(I-K)$ and thus $\operatorname{codim}\operatorname{Im}(I-K) = \operatorname{dim}\operatorname{Ker}(I-K)$?
Comments:
In particular, if $\operatorname{Ker}(I-K)=\{0\}\cong E/\operatorname{Im}(I-K)$ then $\operatorname{Im}(I-K)= E$ and conversely. But in fact, to be more precise, one only has $E/F \cong \operatorname{Ker}(I-K) $...
Just to dissipate possible confusions, the usual isomorphism theorem reads $E/\operatorname{Ker}(I-K) \cong \operatorname{Im}(I-K)$ (but these spaces are infinite dimensional in general).
As we can guess, my statement is probably wrong in general so the question now is to convince ourseves by finding an example of two isomorphic closed subspaces $F_1\cong F_2$ of $E$ such that $E/F_1$ not isomorphic to $E/F_2$. In terms of dimension, it seems plausible that $\operatorname{dim}(E/F_1)="+\infty - \infty" = k_1$ but $\operatorname{dim}(E/F_2)="+\infty - \infty" = k_2 \neq k_1$.
All this eventually amounts to exhibiting a (Fredholm) operator $T$ of non-zero index which is conveniently defined as $\operatorname{dim}(T)=\operatorname{dim}\operatorname{Ker}(T)- \operatorname{codim}\operatorname{Im}(T)$. (Fredholm op. is such that $\operatorname{dim}\operatorname{Ker}(T)< \infty$ and $\operatorname{Im}(T)$ closed so that the quotien $E/\operatorname{Im}(T)$ is Banach)
Next steps of the proof:
$F$ closed as the intersection of kernels of continuous maps. Together with $(I-K)|_F$ is injective, one obtains $\exists\, C\!>0,\, \forall\, \mathbf{x}\in\! F,\, \lVert(I-K)(\mathbf{x})\rVert\geq C \lVert\mathbf{x}\rVert$.
(Same proof by absurd in B. Maurey, Lemme 8.3.2 p.111 and L.C. Evans 2. p.725, original proof in H. Brezis).
The consequence of this inequality is that $\operatorname{Im}(I-K)$ is closed: a converging sequence $(y_n)$ in the image is Cauchy but the inequality implies that the sequence $(x_n)$ of preimage is also Cauchy and thus converges. By continuity $\displaystyle (I-K)\big(\lim_{n\to\infty} x_n \big)= \lim_{n\to\infty} (I-K)(x_n)= \lim_{n\to\infty} y_n =y$. This latter is thus also in the image.
Let us write $T:=I-K$. (cf. notations H. Brezis p.9, orthogonal in the dual) $$\operatorname{Ker}(T^*) =\left\lbrace\lambda\in E^*,\ T^*(\lambda)=\lambda\circ T \equiv 0\right\rbrace = \operatorname{Im}(T)^{\perp} = \{\lambda\in E^*,\, \lambda|_{T(E)}\equiv 0\}$$ Since the image is closed, $\operatorname{Im}( T) =\operatorname{Im}(T)^{\perp\perp}\!= \operatorname{Ker}(T^*)^{\perp}$.
Now comes the more tedious part of showing that $\operatorname{codim}\operatorname{Im}(I-K) = \operatorname{dim}\operatorname{Ker}(I-K)$ which uses the argument that there is no infinite sequences of strictly decreasing closed subsets s.t. sthg (B. Maurey, Lemme 8.3.5 p.112 or H. Brezis (c) p.161-162 or L.C. Evans, 5. p.726). That's the part that already seems to follow from point 2.
