A sketch of the proof can be the following.

Let $G$ be a finite group of non-symplectic automorphisms on a $K3$ surface $X$. Since $G$ is non-symplectic, there exists $g \in G$ such that $g \omega \neq \omega$, where $\omega$ is the holomorphic $2$-form on $X$.

Then, setting $Y:=X/G$, one has $q(Y)=p_g(Y)=0$, since $q(X)=0$ and by the previous remark the holomorphic $2$-form $\omega$ does not descend to the quotient.

From this, one proves that either $Y$ is rational (i.e, bimeromorphic to $\mathbb{P}^2$) or the minimal desingularization of $Y$ is an Enriques surface. In any case, $Y$ is an *algebraic* surface, so there exists an ample divisor $H$ on $Y$.

Finally, the quotient $\pi \colon X \longrightarrow Y=X/G$ is a finite morphism since $G$ is a finite group. It follows that $\pi^*H$ is an ample divisor on $X$, hence $X$ is projective.