$\textbf{All the spaces mentioned in this answer are completely regular}$

I usually call a space $T_{1}$-space ultranormal if whenever there are two disjoint closed sets $A,B\subseteq X$ there is a clopen set with $A\subseteq C,B\subseteq C^{c}$. A strongly zero-dimensional space is usually defined to be a completely regular space $X$ such that the Stone-Cech compactification $\beta X$ is zero-dimensional. Equivalently, a completely regular space $X$ is strongly zero-dimensional if and only if whenever $f:X\rightarrow[0,1]$ is continuous, then there is a clopen set $C$ with $f^{-1}[\{0\}]\subseteq C,f^{-1}[\{1\}]\subseteq C^{c}$. The ultranormal spaces are precisely the normal strongly zero-dimensional spaces, but not every strongly zero-dimensional space is ultranormal. For example, the Tychonoff plank
$(\omega+1)\times(\omega_{1}+1)\setminus\{(\omega,\omega_{1})\}$ is strongly zero-dimensional since $\beta((\omega+1)\times(\omega_{1}+1)\setminus\{(\omega,\omega_{1})\})=(\omega+1)\times(\omega_{1}+1)$ but not ultranormal. I gave a long answer here giving (probably too much) information about the distinction between various notions related to zero-dimensionality.

Let me now prove that every normal strongly zero-dimensional space is ultranormal. Suppose that $X$ is strongly zero-dimensional and normal. Let $C,D\subseteq X$ be disjoint closed sets. Then by normality, there is some continuous $f:X\rightarrow[0,1]$ with $C\subseteq f^{-1}[\{0\}],D\subseteq f^{-1}[\{1\}]$. However, since $X$ is strongly zero-dimensional, there is a clopen set $R$ with $f^{-1}[\{0\}]\subseteq R,f^{-1}[\{1\}]\subseteq R^{c}$. Therefore $C\subseteq f^{-1}[\{0\}],D\subseteq f^{-1}[\{1\}]$.