The answer is yes.

In fact, there is the following result, see Banica- Stanasila, "Algebraic methods in the global theory of Complex Spaces", Theorem 2.12 p. 180.

Theorem Let $f \colon X \to Y$ be a morphism of complex spaces and let $\mathcal{F}$ be a coherent analytic sheaf on $X$, which is flat with respect to $f$. Then the restriction of $f$ to supp($\mathcal{F}$) is an open map.

In particular, every flat morphism is open.

The answer is yes.

In fact, there is the following result, see Banica-Stanasila, Algebraic methods in the global theory of Complex Spaces, Theorem 2.12 p. 180.

Theorem. Let $f \colon X \to Y$ be a morphism of complex spaces and let $\mathscr{F}$ be a coherent analytic sheaf on $X$, which is flat with respect to $f$. Then the restriction of $f$ to supp($\mathscr{F}$) is an open map.

In particular, every flat morphism is open.