I will firstly define the topology $T_{D}$.
Let $D\neq \emptyset$ be a uniform structure on a space $X$.
$U(x)=\{y\in X |(x,y) \in U\}$.
For all $x\in X$, the family $U_{x}=\{U(x) | U \in D\}$ generates the neighborhood system on the point $x$ of a topology on X. Let us denote this topology $T_{D}$.
Claim: If the uniform space $(X,D)$ is Hausdorff, then $(X,T_{D})$ is Hausdorff.
My attempt: Let the uniform space $(X,D)$ be Hausdorff, i.e, $\cap\{U | U\in D\}=\Delta$ (diagonals)
If $x,y\in X$ and $x \neq y$, then $(x,y)\not\in \Delta $ and by the hypothesis, there exists $U\in D$ st. $(x,y)\not\in U$.
Since the symmetric elements of $D$ generates a base, there exists $V\in D$, $V=V^{-1}$ st. $V\circ V \subset U$.
It shows that $V(x)\cap V(y)= \emptyset$, which completes the proof.
Okay, now I want to find an example related to this theorem.
Is it easy to generate a topology satisfying this theorem?
Thank you..
The proof is fine as it is. A uniform spac $(X, \mathcal{D})$ is called "separated" if $\bigcap\{U : U \in \mathcal{D}\} = \Delta_X$. So you're showing that if $(X,\mathcal{D})$ is a separated uniformity then $X$ in the associated topology is Hausdorff.
If $x \neq y$ in $X$ then $(x,y) \notin \Delta_X$ so for some $U_1 \in \mathcal{D}$ we have that $(x,y) \notin U_1$. Let $U_2 \in \mathcal{D}$ be such that $U_2 \circ U_2 \subseteq U_1$, and define $U_3 = U_2 \cap (-U_2) \in \mathcal{D}$. Then indeed $U_3[x] \cap U_3[y] = \emptyset$. For suppose not, and we' d have $z \in U_3[x] ,z \in U_3[y]$ so that $(x,z) \in U_3$, $(y,z) \in U_3$. But then $(x,z) \in U_2, (z,y) \in U_2$ and so $(x,y) \in U_2 \circ U_2 \subseteq U_1$, which is a contradiction with how $U_1$ was chosen.
This situation happens for any uniformity induced by a metric $d$ on $X$ (with base all entourages of the form $U(r) = \{(x,y) \in X^2: d(x,y) < r\}$) as for $x \neq y$ the entourage $D(\frac{d(x,y)}{2})$ does not contain $(x,y)$.
This also happens in uniformities induced by a topological groups structure (the left and right uniformity alike) when we assume that the underlying topological group is at least $T_0$ to begin with.