Explanation of proof that $c_{0}$ doesn't complemented in $l^{\infty}$

Question

I've stuck with a problem. We want to prove that $c_{0}$ doesn't complemented in $l^{\infty}$.

There is a proof given in Complement of $c_{0}$ in $l^{\infty}$.

The main problem I have connected with the second part of proof :

"Now fix $n$ and $k$ such that $I_{n,k}$ is uncountable. Let $J \subset I_{n,k}$ be finite and consider $y = \sum_{j \in J} \operatorname{sign}{[(Px_j)(n)]} \cdot x_j$. Note that $$ (Py)(n) = \sum_{j \in J} \operatorname{sign}{[(Px_j)(n)]}\cdot (Px_j)(n) \geq \frac{\# J}{k} $$ by our choice of $y$. Since $A_i \cap A_j$ is finite for $i \neq j$, we can write $y = f + z$ where $f$ has finite support and $\|z\| \leq 1$. Thus $P(y) = P(f) + P(z) = P(z)$ by hypothesis on $P$ and therefore $\|P(y)\| \leq \|P\| \|z\| \leq \|P\|$. This yields $$\# J \leq \|P\| k$$ whence the absurdity that $I_{n,k}$ must be finite".

As it said we consider finite subset $J \subset I_{n,k}$. And then prove that $\#J \le \|P\| k$, so it shows that $J$ is finite. It confuses me. And even if we show that $J$ is finite, how it shows that $I_{n,k}$ is finite?

Answer 1

The situation is the following: there is an uncountable set $S$ (here $I_{n,k}$). It is proven that there exists a constant $C$ (here $\left\lVert P\right\rVert k$) such that each finite subset of $S$ has at most $C$ elements. This proves that $S$ is finite; otherwise, pick a finite subset with for example $\lfloor C\rfloor +2$ elements.