For a locally compact Hausdorff space $K$, we denote by $C_0(K)$ the Banach space of continuous functions from $K$ to $\mathbb R$ which vanish at infinity equipped with the usual supremum norm.

For any Banach space $X$, a closed subspace $E$ of $X$ is said to be an $M$-summand of $X$ if there exists a subspace $F$ such that $X=E\oplus F$ and for any $x=u+v$, with $u\in E$ and $v \in F$, we have $$ \|x\|=\max\{\|u\|,\|v\|\}. $$ We say that $E$ is a non-trivial M-summand if $\emptyset \subsetneqq E \subsetneqq X$.

Considering the space $C_0(K)$, when $K$ is disconnected, fixing two disjointed non-empty clopens $U, V\subset K$ such that $U\cup V=K$, the closed subspace $$ E_U=\{f\in C_0(K): f|_{U^c}\equiv0\} $$ is a non-trivial $M$-summand of $C_0(K)$, for the closed subspace $$ F_U=\{f\in C_0(K): f|_{V^c}\equiv0\} $$ complements $E_U$ and the condition on the norm is clearly satisfied.

My question is: what about the reverse? That is, if $K$ is connected, is it possible that $C_0(K)$ contains a non-trivial $M$-summand?

I approached this problem as follows. Pick $E$ a non-trivial $M$-summand of $C_0(K)$ and $F$ its complement. Consider the set $$ K_E=\{k\in K | \exists f\in E: |f(k)|>\|f\|/2\}. $$ It surely is an open set of $K$. Picking $K_F$ the similar definition for $F$, it is also an open set of $K$. It is immediate to see that $K_E\cap K_F=\emptyset$, for if $k\in K_E\cap K_F$, then it would exist $f\in K_E$, $g\in K_F$ with $\|f\|=\|g\|=c$ and a scalar $|\lambda|=1$ such that $$ \|f+\lambda g\|=\max\{\|f\|,\|\lambda g\|\}=c, $$ but $$ |f(k)+\lambda g(k)|=|f(k)|+|g(k)|> c/2 + c/2 = c = \|f+\lambda g\|, $$ an absurd.

I want to prove that $K_E$ and $K_F$ are clopens by concluding that $K=K_E\cup K_F$. However, I didn't find a good approach. My intuition says that it is not true, but I have no idea or repertory to conclude so.