A topological space $(X,\mathcal T)$ is said to be divisible iff for each neighborhood $U$ of the diagonal $\Delta=\{(x,x)\mid x\in X\}$ in $X\times X$, there is a symmetric neighborhood $V$ of the diagonal such that: $$V\circ V\subseteq U$$
Is every $T_4$ topological space divisible?
Unfortunately not, but (the way I know how) to demonstrate this is not entirely simple
Definition: A T$_1$-space $X$ is called collectionwise normal if for every discrete family $\{ F_i : i \in I \}$ of closed subsets of $X$ there is a pairwise disjoint family $\{ W_i : i \in I \}$ of open subsets of $X$ such that $F_i \subseteq W_i$ for all $i \in I$.
Theorem (H.J. Cohen, 1952): Every (completely regular) divisible space is collectionwise normal.
proof. Let $\{ F_i : i \in I \}$ be a discrete family of closed subsets of $X$. For each $i \in I$ let $U_i = X \setminus \bigcup_{j \neq i} F_j$, and let $U = \bigcup_{i \in I} ( U_i \times U_i )$. Clearly $U$ is an open neighbourhood of the diagonal $\triangle \subseteq X \times X$, and so by divisibility there is a symmetric neighbourhood $V$ of $\triangle$ such that $V \circ V \subseteq U$. For each $i \in I$ define $$W_i = \bigcup_{y \in F_i} \{ x \in X : \langle x , y \rangle \in V \}.$$ It can be shown that the family $\{ W_i : i \in I \}$ is as required. $\hspace{0.5cm}$$\Box$
(Perhaps for this reason divisible spaces are sometimes called strongly collectionwise normal.)
And there are normal spaces which are not collectionwise normal, for example Bing's $G$.
[Cohen's theorem is from his article Sur un problème de M. Dieudonné, C. R. Acad. Sci. Paris 234, (1952). 290–292.]
Note that being divisible, a.k.a. strongly collectionwise normal, is equivalent to every open cover being uniform, that is, every open cover has a refinement such that if two points are connected by a chain of two sets in the refinement with non-empty intersection (e.g. there are points $x,y,z$ with $x,y$ belonging to some member and $y,z$ belonging to some other member) then these two points must belong to a common member of the original open cover.
The following example of a non-divisible space was constructed (in French) by Cohon in the article linked in user642796's answer.
Consider the product of $\omega_1$ with its order topology and $\omega_1+1$ with its order topology, but with every point of $\omega_1^2$ isolated.
For each point $\alpha<\omega_1$, consider the open set $U_\alpha=(\leftarrow,\alpha]\times (\alpha,\rightarrow)$. These (along with singletons of $\omega_1^2$) form a cover. We will show it is not uniform.
Given some refinement of this cover, each $0<\alpha<\omega_1$ has some $f(\alpha)<\alpha< g(\alpha)$ with $V_\alpha=(f(\alpha),\alpha]\times [g(\alpha),\rightarrow)$ being contained in some member of the refinement. We may apply the pressing down lemma to $f$ to obtain an unbounded (in fact, stationary) subset $S\subseteq \omega_1$ with $f\upharpoonright S$ having constant value $\alpha$. In particular, we have some $g(\alpha+1)\leq\beta<\omega_1$ with $f(\beta)=\alpha$. It follows that $\langle\alpha+1,g(\alpha+1)\rangle,\langle\alpha+1,\omega_1\rangle\in V_{\alpha+1}$, and $\langle\alpha+1,\omega_1\rangle,\langle g(\alpha+1),\omega_1\rangle\in V_\beta$.
However, we note that if $\langle\alpha+1,g(\alpha+1)\rangle$ belongs to some $U_\gamma$ in the original cover, we have $\gamma<g(\alpha+1)$. On the other hand, if $\langle g(\alpha+1),\omega_1\rangle$ belongs to some $U_\gamma$, we have $g(\alpha+1)\leq\gamma$.
Therefore, we conclude that $\langle\alpha+1,g(\alpha+1)\rangle$ and $\langle g(\alpha+1),\omega_1\rangle$ do not belong to any common member of the original cover, despite being chained by a pair of members $V_{\alpha+1},V_\beta$ of the arbitrary refinement. This proves that the constructed cover is not uniform.
To see that this is your desired counter-example, let's first observe that the space is $T_2$. Then to conclude, "we'll easily see" (to quote Cohen) that this space is in fact collectionwise normal.
The first step is to observe that for a discrete collection of closed sets $\{F_i:i\in I\}$, we may note that $\{F_i\cap(\omega_1\times\{\omega_1\}):i\in I\}$ is a discrete collection of closed sets in the subspace $\omega_1\times\{\omega_1\}$. This is a copy of the linearly ordered topological space (LOTS) $\omega_1$, and every LOTS is collectionwise normal. So we have pairwise disjoint open subsets $U_i\subseteq\omega_1$ with $F_i\cap(\omega_1\times\{\omega_1\})\subseteq U_i\times\{\omega_1\}$.
Then we note that for each $i\in I$, there must be some $\alpha_i<\omega_1$ with $F_i\cap(\omega_1\times(\alpha_i,\rightarrow))\subseteq F_i\cap(U_i\times(\alpha_i,\rightarrow))$. Finally, we note that $V_i=\big(U_i\times(\alpha_i,\rightarrow)\big) \cup \big(F_i\cap(\omega_1\times(\leftarrow,\alpha_1])\big)$ is a pairwise disjoint family of open subsets of our space with $F_i\subseteq V_i$. (Edit: actually I think this isn't quite right; more work is required to ensure some other $F_j$ doesn't "reach out" into $F_i$'s space, but I don't have a clever argument to fix it this morning.)
