It is well known that every topological space is quasi-uniformizable, and every quasi-uniform space $(X, \mathcal U)$ induces a natural topology on $X$. Moreover, if $(X, \mathcal U)$ is a quasi-uniform space, then there is a family of quasi-pseudometrics $\{p\_{\alpha}\}\_{\alpha \in A}$ which generates the quasi-uniformity on $X$, in the sense that the sets $$V\_{\alpha, \varepsilon} := \{(x,y) \in X^2 | p\_{\alpha}(x,y) < \varepsilon\}$$ form a base for the quasi-uniformity $\mathcal U$ (one can also replace $<$ by $\leq$ in the definition of the $V\_{\alpha, \varepsilon}$). Moreover, these quasi-pseudometrics are quasi-uniformly continuous, in the sense that to every $\varepsilon>0$, there is $U \in \mathcal U$ such that if $((x,y), (x', y') ) \in U \times U$, then $|p\_{\alpha}(x,y) - p\_{\alpha}(x', y') | < \varepsilon$. Further, if $\mathcal U$ is a uniformity, it is possible to take the $p\_{\alpha}$ to be pseudometrics.

As per usual, when speaking of the topology on a (quasi)uniform space, the canonical topology generated by the (quasi)uniformity will always be meant. 

Every uniform space is completely regular (hence regular); this is not true for quasi-uniform spaces by the preliminary remarks above. One way to show uniform spaces are regular would be as follows: 

Let $(X, \mathcal U)$ be a uniform space. Take a family of uniformly continuous (in the sense described above) pseudometrics $\{p\_{\alpha}\}\_{\alpha}$ generating the uniformity. The sets $$F\_{\alpha, \varepsilon}: = \{(x,y) \in X^2 | p\_{\alpha}(x,y) \leq \varepsilon\}$$ form a base for the uniformity. These sets are closed in the product topology of $X \times X$, because they are the preimages of $[0, \varepsilon]$ under the maps $X \times X \rightarrow [0, \infty): (x,y) \mapsto p\_{\alpha}(x,y)$, which are uniformly continuous (hence continuous). So for every $x \in X$, $\{F\_{\alpha, \varepsilon}[x] \}\_{\alpha, \varepsilon}$ forms a local base at $x$ consisting of closed sets. Regularity follows. 

My question: where does this argument fail in the case of quasi-uniform spaces? 

I suspect the argument fails because the following may not hold in quasi-uniform spaces (I do not have a counterexample, though): if $V$ is an entourage which is closed in the product topology $X \times X$, then $V[x] $ is a closed subset of $X$ for every $x \in X$. In uniform spaces, one way to show this is to use the fact that if $E \subseteq X \times X$, then $$\overline E = \bigcap\_{U~ \text{symmetric entourage}} U \circ E \circ U.$$ Since $U^{-1}$ need not be an entourage whenever $U$ is an entourage in a quasi-uniformity, I do have my suspicions about $V[x]$ being closed whenever $V \in \mathcal U$ is closed in $X \times X$. Besides this point, I do not know where else the argument goes wrong, since any quasi-uniformly continuous map between quasi-uniform spaces is continuous in the induced topologies.

Thanks in advance for reading my question. If I need to clarify anything or add more details, do let me know.