I found a proof that the existence of a Schauder basis implies the bounded approximation property on Banach spaces. ( J.Lindenstrauss, L.Tzafriri; classical Banach spaces; p.12).
Let $\{P_i\}$ be the projections associated to the basis $\{x_n\}$ and $A = \sup\limits_i ||P_i||$. For every compact set of vectors $K\subset X$ and every $\varepsilon >0$ we can find $i = i(\varepsilon,K)$ such that the distance $$d(x, span\{x_1, ... , x_i\}) < \varepsilon / 2A; x\in K $$ Then obviously, $||x-P_ix|| < \varepsilon $, for every $x\in K$
I don't understand why hes using $\varepsilon/2A$ as the bound for the distance. Wouldn't $\varepsilon$ be sufficient because $||x-P_ix|| = d(x, span\{x_1, ... , x_i\})$?
Essentially this is saying that the $\varepsilon$ depends on $A:=\sup_i \|P_i\|$.
Since $\varepsilon$ is arbitrary we can, of course, just write $\varepsilon$ in the inequality and know that we are correct. This hides the dependency on $A$ though, and for subsequent applications that can be a bad thing. Note that the number of vectors you need to take in the linear span also depends on this $\varepsilon$. If $A$ is very close to $1$ then you will need (many) more vectors in your span than if $A\approx 10^5$ (for example).