As indicated by Theo Buehler above, Fremlin proves what I want and much more in his book. However, the proof given in this reference can be simplified a lot in my setting, so that I can answer my own question by giving a relatively simple proof :
Basically, Fremlin's proof begins by finding a projective resolution of $K$ inside the dual ball of $C(K)$. But the existence of a projective resolution in $\mathbf{CHaus}$ has a short proof (see this article), so I use it freely below.
Proof of the direction "$\leftarrow$" in $\star$ : Let $X$ be a projective resolution of $K$ in $\mathbf{CHaus}$, and let $p : X \twoheadrightarrow K$ be the corresponding map, which satisfies $p(S) \neq K$ whenever $S$ is a proper closed subspace of $X$. Then $p$ induce an isometry $\tilde p : f \in C(K) \rightarrow f \circ p \in C(X)$. Since $\mathrm{Im} \; \tilde p$ is isometric to $C(K)$, which has the Hahn-Banach extension property, we can extend the identity on $\mathrm{Im} \; \tilde p$ to a norm-one operator $T : C(X) \rightarrow \mathrm{Im} \; \tilde p$.
Let $h$ be in the unit ball of $C(X)$, and set $S=\overline{ \lbrace h \neq 0 \rbrace } \cup \{ Th = 0 \}$. If $S$ is a proper subspace of $X$ then some $p(x) \in K$ doesn't belong to $p(S)$. Let $f \in C(K,\[ 0,1 \])$ be such that $f \circ p(x) = 1$ and $f_{|p(S)}=0$. Then $\lVert \tilde p (f) \pm h \rVert \leq 1$, so that $\lVert \tilde p (f) \pm Th \rVert = \lVert T( \tilde p (f) \pm h )\rVert \leq 1$. But for an appropiate choice of sign, the value of $\tilde p (f) \pm Th$ at $x$ exceeds $1$, a contradiction. Thus $S=X$.
Let $x_1$ and $x_2$ be distinct points in $X$, hence separated by disjoint closed sets $F_1$ and $F_2$ in $X$. Let $h \in C(X,\[ 0,1 \])$ be such that $h_{|F_1}=0$ and $h_{|F_2}=1$. Then $\overline{ \lbrace h \neq 0 \rbrace } \subset {}^c( \mathring F_1)$, so that by the point above $x_1 \in \mathring F_1 \subset \lbrace Th = 0 \rbrace$, and $Th(x_1)=0$. Similarly, $Th(x_2)=1$. Since $Th \in \mathrm{Im} \; \tilde p$, this shows that $p(x_1) \neq p(x_2)$.
We have shown that $p$ is injective, so that $p$ is a homeomorphism $X \simeq K$. Thus $K$ is projective.
