If $X,Y,Z$ are Banach spaces and $u: X \to Y. v: Y \to Z$ are Fredholm then $\mathrm{ker}(vu)$ is finite dimensional.
Can one argue as follows?:
If $x \in \mathrm{ker}(vu)$ then either $x \in \mathrm{ker}(u)$ or $u(x) \in \mathrm{ker}(v)$. Hence $$ \mathrm{dim}(\mathrm{ker(vu)}) \le \mathrm{dim}(\mathrm{ker(u)}) + \mathrm{dim}(\mathrm{ker(v)})$$
It seems to be intuitively clear but since I don't have a formal proof for the $\le$ above I am skeptical.
I'll prove slightly more
Consider commutative diagramm $$ \require{AMScd} \begin{CD} 0 @>>> X @>(\begin{smallmatrix} 1_X\\ u\end{smallmatrix})>> X\oplus Y @>\begin{pmatrix} -u\; 1_Y\end{pmatrix}>> Y @>>> 0\\ @. @VVuV @VVvu\oplus 1_YV @VVvV @.\\ 0 @>>> Y @>>(\begin{smallmatrix} v\\ 1_Y\end{smallmatrix})> Z\oplus Y @>>\begin{pmatrix} -1_Z\; v\end{pmatrix}> Z @>>> 0\\ \end{CD} $$ then by snake lemma we have the following exact sequence $$ 0\to\operatorname{ker}(u)\to\operatorname{ker}(vu)\to\operatorname{ker}(v)\to\operatorname{coker}(u)\to\operatorname{coker}(vu)\to\operatorname{coker}(v)\to 0 \tag{1} $$ Since $\operatorname{ker}(u)$, $\operatorname{coker}(u)$, $\operatorname{ker}(v)$, $\operatorname{coker}(v)$ are finite dimensional and the sequence $(1)$ is exact then $\operatorname{ker}(vu)$ and $\operatorname{coker}(vu)$ are finite dimensional too.
From rank nullity theorem it easily follows that for any exact sequence of finite dimensional vector spaces $$ 0\to V_0\to V_1\to\ldots\to V_{n-1}\to V_n\to 0 $$ holds $$ \sum_{i=0}^n (-1)^i \operatorname{dim}(V_i)=0 $$ Applying this theorem to $(1)$ we get $$ \operatorname{dim}\operatorname{ker}(u)- \operatorname{dim}\operatorname{ker}(vu)+ \operatorname{dim}\operatorname{ker}(v)- \operatorname{dim}\operatorname{coker}(u)+ \operatorname{dim}\operatorname{coker}(vu)- \operatorname{dim}\operatorname{coker}(v)=0 $$ Recall the definition of index $$ \operatorname{ind}(w)=\operatorname{dim}\operatorname{ker}(w)-\operatorname{dim}\operatorname{coker}(w) $$ to get $$ \operatorname{ind}(v)+\operatorname{ind}(u)-\operatorname{ind}(vu)=0 $$ The last is well known index theorem for Fredholm operators.
While my question is still open and I'm still wondering whether I can argue as I did in my question I came up with another way of proving it:
Note that $$0 \to \mathrm{ker}(u) \xrightarrow{i} \mathrm{ker}(vu) \xrightarrow{u}\mathrm{ker}(v) \cap u(X) \xrightarrow{v}0$$
is a short exact sequence such that $i$ has a one sided inverse. Hence by the splitting lemma $$ \mathrm{ker}(vu) = \mathrm{ker}(u) \oplus \mathrm{ker}(v) \cap u(X)$$
hence $\mathrm{ker}(vu) $ is finite dimensional.
