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I think the answer is yes. to see the sketch of proof you could suppose on the contrary that, $f$ is not bounded and choose $$x_1, x_2,...,x_m,... \in X$$ and$$r_1,r_2,...r_m,...\in \mathbb{R}^+$$ so that $$r_1< |f(x_1)|< r_2<|f(x_2)|< ... r_m< |f(x_m)|< ...$$ Because $X$ is zero-dimensional and $f$ is continuous, for each $m\in \mathbb{N}$ you could find a clopen subset $V_m \subseteq${$x\in X: r_m< |f(x)|< r_{m+1}$}. You could see that for $i\neq j, V_i\cap V_j=\emptyset$. Moreover $$V_0=X-\cup_{i=1}^{\infty}V_{i} $$ which is equal to $\cup_{i=1}^{\infty}(${$x\in X: |f(x)|< r_{i+1}$}$-V_1\cup V_2 \cup... \cup V_i$$)$, and obviously it is open. It yields that $X$ has an infinite partition and this implies a contradiction. because $X$ is $\omega-$ pseudocompact and has no countable partition of clopen sets.