Is this proof that all metric spaces are Hausdorff spaces correct? Let $x$ and $y$ be distinct points of a metric space $M$. Prove that there exist in $M$ disjoint open sets $U$ and $V$ with $x \in U$ and $y \in V$.
Let $U$ and $V$ be open balls centered at $a$ and $b$, respectively. Also, $a, b \in M, x \in U, y \in V$. Since $M$ is a metric space, $M$ has a real defined distance function $D(x,y) < \epsilon$. Since $x\ne y$, $y$ is a limit point of $U$ for $D(x,y)<\epsilon.$ That is, $B_{r}(x)\subset U.$ Thus, an open set is contained in $U$, since open balls are open sets.
Similarly, $x$ is a limit point of $V$. Applying the same definitions as we did to $U$, we find an open set in $V$. These open sets are disjoint. QED.
 A: 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...
A: 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.
