Sobcyzk's Theorem

Motivation:
In a Banach space $X$, $M \leq X$ is complemented if there exists $N \leq X$ such that $N + M = X$, $N \cap M = 0$.
We showed $M$ is complemented iff there is a bounded projection $P:X \rightarrow X$ such that $\textrm{ran }P = M$.

Remark: $c_0$ is not complemented in $\ell^\infty$.

Theorem 3.1.1 (Sobcyzk, 1941)  
Let $Y \leq X$, $X$ a separable Banach space such that $Y$ is isometrically isomorphic to $c_0$. Then, there is a projection of $X$ onto $Y$ of norm at most $2$.

It suffices to prove that if $T_0 \in B(Y,c_0)$ with $\Vert T_0\Vert = \lambda$, then there exists $T \in B(X,c_0)$ with $\Vert T\Vert \leq 2 \lambda$ (apply this result with $T_0$ the identity on $Y$).

Proof. (Veech, PAMS, '71)
Let $S = \overline{B_\lambda(0)}$ in $X^*$ with the wk $^*$-topology. We know $S$ is compact and metrizable. Let $d$ be a metric giving this topology. Notice $T_0(Y)$ are sequences.
For each $n \in \mathbb{N}$, by Hahn-Banach, there is $\phi_n \in X^*$ so that $\phi_n(y)$ is the $n^{th}$ element of $T_0(y)$ for all $y \in Y$.

Let $K = S \cap Y^\perp$.

Claim: Every wk $^*$-cluster point of $(\phi_n)$ is in $K$.
If $\phi_{n_k} \rightarrow \phi$ in wk $^*$, then

$$\phi(y) = \lim_k \phi_{n_k}(y) = \lim_k (T_0(y))_{n_k} = 0$$

as $T_0(y) \in c_0$. That is, dist $(\phi_n,K) \rightarrow 0$ as $n \rightarrow \infty$. Pick $\psi_n \in K$ such that $\Vert\phi_n - \psi_n\Vert =$   dist $( \phi_n,K )$ (as $K$ is wk $^*$-compact and dist  is lower semicontinuous). Thus, $)$ is the only wk $^*$-cluster point of $(\phi_n - \psi_n)$ in $\overline{B_{2\lambda}(0)}$ in $X^*$.
Since this sequence has a wk $^*$-cluster point, we have

$$(\phi_n - \psi_n) \rightarrow 0$$

in the wk $^*$-topology.
Define $T:X \rightarrow c_0$ by

$$(T(x))_n = \phi_n(x) - \psi_n(x).$$

Clearly, $T$ extends $T_0$. ( $\phi_n \subset Y^\perp$) and $\Vert T\Vert \leq 2 \lambda$ ( $\phi_n - \psi_n \in \overline{B_{2\lambda}(0)}$). Then $T$ is the required projection if $T_0$ is the identity map. $\qedsymbol$