Relationship between $c_0$ and $C(K)$ Suppose that $K$ is an infinite compact metric space. Define $c_0=\{ (x_n)_{n \in \mathbb{N}}| \lim_n\| x_n \|=0 \}$ and $C(K)$ the set of real-valued continuous functions on $K$.
Question: Is $C(K) \subseteq c_0$? In this paper, in the proof of Theorem $5$, there is this sentence:

Since the spaces $C(K)$ are absolute Lipschitz retracts, there is a Lipschitz retraction from $c_0$ onto $C(K)$ which maps $0$ to $0$.

From what I understand from the statement above, it seems that $C(K)$ is a subset of $c_0$. Otherwise there won't be any retraction from $C(K)$ onto $c_0$.
Remark: A Banach space is called absolute Lipschitz retract if for every metric space $Y$ containing $X$, there exists a retraction $r:X \rightarrow X$ such that $r$ is also Lipschitz. 
 A: The answer to your question is included in the third line of the proof of the theorem you are interested in!

By a classical result of Aharoni (see Theorem 7.11, p. 176 in 1), we know that there is a 3-Lipschitz-homeomorphism between $C(K)$ and some subset of $c_0$.

This proof can be summarized in four steps as follows:


*

*By Sobczyk's theorem, any (linear) copy of $c_0$ in $C(K)$ is 2-complemented. (See my answer in this thread.)

*By Aharoni's theorem $C(K)$ embeds into $c_0$ by a 3-Lipschitz homeomorphism (this time this map cannot be expected to be linear as there are obstacles concerning the Szlenk index) and we get actually a retraction from $c_0$ onto the range of this embedding.

*We lift these morphisms to linear maps between the free spaces which give linear embeddings with complemented ranges.

*We apply the Pełczyński decomposition method to get an isomorphism between free spaces.
A: That proof starts with the sentence that $c_0$ is $2$-complemented in $C(K)$. And indeed $c_0$ is isometrically embedded into $C(K)$. This holds as $K$ is an infinite compact metric space, so has a convergent sequence (with all different terms) $x_n$ with limit $x$ (unequal to all $x_n$). Call $S = \{x_n: n \in \mathbb{N}\} \cup \{x\}$. Then $C(S)$ embeds into $C(K)$ (this is due to Dugundji, I think) and there is a projection of $C(K)$ onto $C(S)$ (restriction), and $\{f \in C(S): f(x) = 0\}$ is isometric to $c_0$.  
