A type $\Phi$ in a complete theory corresponds to a closed subset $[\Phi]$ of the space of types $S_n(X)$.
According to your definition, a type is algebraic if the closed subset defined by it is covered by a family of open subsets $[\varphi]$ corresponding to algebraic formulas $\varphi$.
But the Stone space is compact, and the type is a closed subset, so $[\Phi]$ is covered by finitely many of these, so there are algebraic $\varphi_1,\ldots,\varphi_n$ such that $[\Phi]\subseteq \bigcup_{j=1}^n [\varphi_j]$, but the latter is just equal to $[\bigvee_{j=1}^n \varphi_j]$, so $\Phi\vdash \bigvee_{j=1}^n \varphi_j$, and $\bigvee_{j=1}^n \varphi_j$ is of course algebraic.