Suppose every complete subset of a uniform space is closed. Is it Hausdorff?

Question

Suppose

• $(X,\mathcal{D})$ is a uniform space.
• for each nonempty subset $A\subseteq X$, if the subspace $(A,\mathcal{D}_A)$ is complete then $A$ is closed.

Is $(X,\mathcal{D})$ Hausdorff?

Answer 1

First, notice that $X$ is $T_0$: the uniform structure induced on a singleton make it a complete uniform space, so any singleton is closed. Then, Hausdorffness follows from the results:

Theorem 1: A topological space is uniformizable iff it is $T_{3 \frac{1}{2}}$.

Theorem 2: $T_0+ T_{3 \frac{1}{2}}$ is equivalent to $T_2+T_{3 \frac{1}{2}}$.

I can add some indications if needed. For example, the second theorem results from a construction which showed that any $T_0+T_{3 \frac{1}{2}}$ topological space is homeomorphic to a subspace of some cube $[0,1]^I$.