Let $E$ be a Banach space. Let $\mathcal K(E)$ be the space of all compact (bounded linear) operators from $E$ to $E$. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. Let $I:E \to E$ be the identity map.
I'm reading the proof of part (d) of the following theorem in Brezis' Functional Analysis.
Theorem Let $T \in \mathcal K(E)$. Then
(a) $\dim N(I-T) < \infty$,
(b) $R(I-T)$ is closed, and more precisely $R(I-T) = N(I-T^*)^\perp$,
(c) $N(I-T) = \{0\} \iff R(I-T) = E$,
(d) $\dim N(I-T)=\dim N(I-T^*)$.
Could you explain how we get a contradiction in the third paragraph below?
Proof of (d) Let $d:= \dim N(I-T)$ and $d^* :=\dim N(I-T^*)$. First, we will prove $d^* \le d$. Assume the contrary that $d < d^*$. By (a), $N(I-T)$ has a complement in $E$. Then there is a (continuous linear) projection map $P:E \to N(I-T)$ which is surjective. By (b), $\operatorname{codim} R(I-T)=d^*$. Then has a complement $F$ of dimension $d^*$. Since $d < d^*$, there is an injective linear map $\Lambda : N(I-T) \to F$ which is not surjective. Let $S:= T + \Lambda P$. Then $S \in \mathcal K(E)$ since $\Lambda P$ has finite-rank.
We claim that $N(I-S) = \{0\}$. Indeed, if $0=(I-S)u$, then $(I-T)u = \Lambda P u$. Because $R(I-T) \cap R(\Lambda) \subset R(I-T) \cap F = \{0\}$, we get $u \in N(I-T)$ and thus $Pu=u$. Then $0 = \Lambda u$. Because $\Lambda$ is injective, we get $u=0$.
Applying (c) to $S$, we get $R(I-S)= E$. This is absurd, since there is some $f \in F \setminus R(\Lambda)$, and so the equation $u-Su=f$ has no solution.
Hence we have proved that $d^* \le d$. Applying this fact to $T^*$, we have $$ \dim N(I-T^{**}) \le \dim N(I-T^*) \le \dim N(I-T). $$
But $N(I-T^{**}) \supset N(I-T)$ and therefore $d=d^*$.
$(u-Su) = f \implies (I-T)u - \Lambda Pu = f$
$(I-T)u \in R(I-T)$ and $f,\Lambda Pu \in R(I-T)^\perp$. So $(I-T)u = 0$.
Hence $\Lambda Pu = -f$, but $f \notin R(\Lambda)$. So $u - Su = f$ has no solution.