Any uniformity on a set $X$ is generated by a family of pseudometrics on $X$. So if $(X,\mathcal D)$ is a uniform space there's a set $P$ of pseudometrics on $X$ with

$$\mathcal D=\left< \bigcup_{d\in P} \mathcal D_d\right> \tag1 $$

where the angles show the filter generated by the subbace $\bigcup_{d\in P} \mathcal D_d$; where each $\mathcal D_d$ is the uniformity on $X$ with base $$\lbrace U_d(r) \mid r>0 \rbrace$$ in which $$U_d(r)=\lbrace (x,y)\in X^2\mid d(x,y)\le r\rbrace$$

$(1)$ immediately implies that $(X,\mathcal D)$ can be embedded uniformly in a product of pseudometric spaces.

Now suppose $(X,\mathcal D)$ is Hausdorff. Isbell says $(X,\mathcal D)$ can be embedded in a product of metric spaces. (The proof given uses covering definition of uniform space, while I'm only familiar with diagonal definition).

My question is:

Is any Hausdorff uniformity on a set $X$ generated by a family of metrics on $X$? Is there a couterexample?

I think none of the pseudometrics that generate a Hausdorff uniformity need to be a metric. Because: $$\bigcap_{d\in P}\bigcap_{n\in \Bbb N}U_d(\frac{1}{n})=\Delta_X \text{ (diagonal)}$$ but this intersection need not be countable and so there may be no metrics in the set $P$ (of pseudometrics that generate the Hausdorff uniformity $\mathcal D$). So uniformities generated by metrics need not be all Hausdorff uniformities.