For this question, we say that a zero-dimensional space $X$ is $\omega$-pseudocompact if every partition of $X$ into clopen sets is finite. In other words, a zero-dimensional space $X$ is $\omega$-pseudocompact if and only if every continuous mapping $f:X\rightarrow\mathbb{N}$ is bounded.
One can easily show that a zero-dimensional space $X$ is $\omega$-pseudocompact if and only if every countable clopen cover of $X$ has a finite subcover. For zero-dimensional spaces, is $\omega$-pseudocompactness equivalent to pseudocompactness?


For strongly zero-dimensional spaces, these properties are equivalent and I give a proof right here. Recall that a completely regular space $X$ is strongly zero-dimensional if the Stone-Cech compactification $\beta X$ is zero-dimensional. It is well known that a space $X$ is strongly zero-dimensional if and only if whenever $Z\_{1},Z\_{2}$ are disjoint zero-sets, then there is a clopen set containing $Z\_{1}$ and disjoint from $Z\_{2}$. If $X$ is strongly zero-dimensional and $f:X\rightarrow[0,\infty)$ is an unbounded continuous mapping, then since $f^{-1}[0,n],f^{-1}[n+1,\infty)$ are disjoint zero-sets, there is a clopen set $A\_{n}$ containing $f^{-1}[0,n]$ but disjoint from $f^{-1}[n+1,\infty)$. The clopen cover $\{A\_{n}|n\in\mathbb{N}\}$ has no finite subcover since $A\_{0}\cup...\cup A\_{n}$ does not intersect $f^{-1}[n+1,\infty]$ for all $n$. Therefore every strongly zero-dimensional space that is not pseudocompact is not $\omega$-pseudocompact, and clearly every pseudocompact space is $\omega$-pseudocompact.



Is every zero-dimensional space with no infinite clopen partition pseudocompact?
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. 


