Let $(X,\mathcal D)$ be a uniform space and $A,B\subseteq X$. Let's say $A$ is *uniformly inside* $B$ and write $A\le B$ iff there's some entourage $D$ for which
$$(\forall a\in A)(D[a]\subseteq B)$$


A uniform space $(X,\mathcal D)$ is *uniformly normal* iff for each open set $U$ and any closed set $C\subseteq U$, there's some open set $W$ such that 
$$C\le W\subseteq \overline W \le U$$


Do you have an example of a normal uniform space that is not uniformly normal?!



Is there a normal space that is not uniformly normal
Yes, since you're using a bad definition. $\;\;\;$ Let $\:\langle X\hspace{.02 in},\hspace{-0.03 in}\mathcal{D}\rangle\:$ be the Euclidean plane with its
  

usual uniform structure. $\;\;\;$ Let $C$ be the closed lower half-plane, and let $U$ be the subset of
  

the plane that is below the graph of $\:x\mapsto \frac1{1+x^{\hspace{.02 in}2}}\;$. $\;\;\;$ $U$ is open and $C$ is closed and $\:C\subseteq U\;$.
  

For any open set $W$, if $\:C\leq W\:$ then there exists a positive $\epsilon$ such that $W$ contains the line $\:y=\epsilon\;$.
  

Since no such line is a subset of $U$, there is no open set $W$ such that $\: C\leq W\subseteq \overline{W} \leq U \;$.
  

Therefore $\:\langle X\hspace{.02 in},\hspace{-0.03 in}\mathcal{D}\rangle\:$ does not satisfy your definition of being uniformly normal.


