Is this proof that all metric spaces are Hausdorff spaces correct?

I think you've made it too complicated. There's no reason to mention limit points.

You want to show that for points $x\ne y$, there are disjoint open sets containing them.

Just use the open neighborhoods of radius $r=D(x,y)/2$ centered at $x$ and $y$.

Since $D(x,y)>0$, then points $x$ and $y$ are members of those respective open neighborhoods.

To show that they are disjoint, suppose $z$ is a point in their intersection. Then $D(x,z)<r$ and $D(y,z)<r$. So by the triangle inequality, $D(x,y)\le D(x,z)+D(z,y)<r+r = D(x,y)$, and we have a contradiction.

As T. Bongers points out, you can't assume that $U$ and $V$ exist in your proof. You can ask yourself what properties $U$ and $V$ must have if they are to satisfy the requirements of the problem statement, and then try to work backwards. But after you figure out a proof strategy, you should start over from the beginning and prove for yourself that $U$ and $V$ exist.

The easiest way to show that $U$ and $V$ exist is to explicitly define them! To produce a definition, it is not enough just to say that they are sets. You have to define them unambiguously, in such a way that anyone can read the definition and understand which points are contained in the sets.

Hint: You can make $U$ an open ball centered and $x$ and $V$ and open ball centered at $y$. To complete the definition, you'll have to define the radius of these balls. What radius should you choose? It will depend somehow on $x$ and $y$. As Sigur suggests, try drawing a picture...

Once you've chosen a definition for $U$ and $V$, it remains only to prove that they are disjoint. This will follow from certain basic facts about metric spaces...