Hausdorff iff convergent filterbase has a unique point of convergence. Is this correct? Suppose $(X, \mathcal{T})$ is a Hausdorff space. Consider a filterbase $\mathcal{B}$ in $X$ such that $\mathcal{B}$ converges to $a\in X$.We show that, if $b \in X$ and $a\neq b$, then $\mathcal{B}$ cannot converge to $b$. Since, $a\neq b$ and $X$ is Hausdorff, there exist open neighbourhoods $U(a) \in \mathcal{T}$ and $V(b) \in \mathcal{T}$ such that $U(a) \cap V(b) =\emptyset$.  Since $\mathcal{B}$ converges to $a$, there exists a filterbasis element $B_j\in \mathcal{B}$ for every open neighbourhood $U(a)$ of $a$ such that $B_j \subset U(a)$. Consequently, $B_j \cap V(b) =\emptyset$. Therefore, $\mathcal{B}$ cannot converge to $b$. [\textit{If it did, there would exist $B_k\in \mathcal{B}$ such that $B_k\subset V(b)$, as a consequence of which $B_j\cap V(b) \neq \emptyset$}].\ Conversely, suppose $X$ is not Hausdorff. We shall construct a filterbase $\mathcal{B}$ that converges to more than one point in $X$. Consider $a$ and $b$ in $X$, since $X$ is not Hausdorff $U(a) \cap V(b) \neq \emptyset$ for all open neighbourhoods $U$ of $a$ and $V$ of $b$. \ Define $\mathcal{B}= \{B_i \mid B_i=U_i \cap V_i; i \in I\}$ where $U$ and $V$ are open neighbourhoods of two distinct points $x$ and $y$ respectively. Clearly, $B_i \neq \emptyset$ for all $B_i \in \mathcal{B}$ and since we can choose $U=X$ for any $i, j \in I$ we can find $k \in I$ such that $B_k \subset B_i \cap B_j = X \cap V_i(y) \cap V_j(y)$. Thus, $\mathcal{B}$ is a filterbase. Every $B_i \in \mathcal{B}$ is such that $B_i = U_i(x) \cap V_i(y)$ and so for every $U(x) and V(y)$ we have a $B_i \in \mathcal{B}$ such that $B_i \subset U(x)$ and $B_i \subset V(y)$. Therefore, $\mathcal{B}$ converges to both $x$ and $y$ in $X$.   
 A: There are a few problems, mostly minor.

Since, $a\neq b$ and $X$ is Hausdorff, there exist open neighbourhoods $U(a) \in \mathcal{T}$ and $V(b) \in \mathcal{T}$ such that $U(a) \cap V(b) =\emptyset$.  

Although your notation $U(a)$ suggests that $U(a)$ is an open nbhd of $a$, and similarly for $V(b)$ and $b$, you never actually said so, and you really should. And since you’re already distinguishing these as $U$ and $V$, there’s no real need for the extra notational complication of $U(a)$ and $V(b)$.

Since $a\ne b$ and $X$ is Hausdorff, there are $U,B\in\mathcal{T}$ such that $a\in U$, $b\in V$ and $U\cap V=\varnothing$.

The rest of this direction of the proof is okay, though you could stand to make the contradiction in the parenthetical If it did remark at the end a bit more explicit:

... as a consequence of which $B_j\cap V\ne\varnothing$, since $B_j\cap B_k\ne\varnothing$ by the definition of a filterbase.

And you could organize it a bit more tightly without changing the idea at all:

Since $\mathcal{B}$ converges to $a$, there is a filterbasis element $B_a\in\mathcal{B}$ such that $B_a\subseteq U$. If $\mathcal{B}$ also converged to $b$, there would be a $B_b\in\mathcal{B}$ such that $B_b\subseteq V$. But then $B_a\cap B_b\subseteq U\cap V=\varnothing$, contradicting the definition of a filterbase. Thus, $\mathcal{B}$ cannot converge to $B$.

In the other direction there’s a mistake near the beginning. You have:

Consider $a$ and $b$ in $X$, since $X$ is not Hausdorff $U(a) \cap V(b) \neq \emptyset$ for all open neighbourhoods $U$ of $a$ and $V$ of $b$.

First, you can’t let $a$ and $b$ be arbitrary points of $X$: there might be only two points in $X$ that can’t separated by disjoint open sets, as is the case, for instance, with the line with two origins. You must specify that you’re starting with points $a$ and $b$ that can’t separated by disjoint open sets. Finally, as a minor point, $U(a)$ and $U$ are not the same thing, so you can’t switch back and forth between them as if they were the same.

Since $X$ is not Hausdorff, there are distinct points $a,b\in X$ such that $U\cap V\ne\varnothing$ whenever $U$ is an open nbhd of $a$ and $V$ is an open nbhd of $b$.

Then you say:

Define $\mathcal{B}= \{B_i \mid B_i=U_i \cap V_i; i \in I\}$ where $U$ and $V$ are open neighbourhoods of two distinct points $x$ and $y$ respectively. 

Where did $x$ and $y$ come from? You’ve not mentioned them before: they come out of nowhere. You should be talking about the specific two points $a$ and $b$ that cannot be separated by disjoint open sets, not about some new points $x$ and $y$. You also haven’t defined the index set $I$. You want $\mathcal{B}$ to be the set of all intersections of an open nbhd of $a$ with an open nbhd of $b$, so just say so:

Let $\mathcal{B}=\{U\cap V:U\text{ is an open nbhd of }a\text{ and }b\text{ is an open nbhd of }b\}$.

If you want to get rid of some of the words, you can do so like this:

Let $\mathcal{N}(a)=\{U\in\mathcal{T}:a\in U\}$, the family of open nbhds of $a$, and let $\mathcal{N}(b)=\{U\in\mathcal{T}:b\in U\}$. Let $\mathcal{B}=\{U\cap V:U\in\mathcal{N}(a)\text{ and }V\in\mathcal{N}(b)\}$.

Don’t bother to index $\mathcal{B}$: there’s no need to do so, and the only really convenient index set is $\mathcal{N}(a)\times\mathcal{N}(b)$, which gives you moderately ugly indices. With or without indices, however, you’re quite correct that it’s clear that each $B\in\mathcal{B}$ is non-empty. The next step, though, has problems. You wrote:

Clearly, $B_i \neq \emptyset$ for all $B_i \in \mathcal{B}$ and since we can choose $U=X$ for any $i, j \in I$ we can find $k \in I$ such that $B_k \subset B_i \cap B_j = X \cap V_i(y) \cap V_j(y)$. 

You have to prove that for any $B,B\,'\in\mathcal{B}$ there is a $B\,''\in\mathcal{B}$ such that $B\,''\subseteq B\cap B\,'$. This means that you can’t specify what open nbhds of $a$ and $b$ are used in forming $B$ and $B\,'$.

Now let $B,B\,'\in\mathcal{B}$ be arbitrary. Then there are open nbhds $U,U'$ of $a$ and $V,V\,'$ of $b$ such that $B=U\cap V$ and $B\,'=U'\cap V\,'$. Clearly $U\cap U'$ is an open nbhd of $a$, and $V\cap V\,'$ is an open nbhd of $b$, so by definition $(U\cap U')\cap(V\cap V\,')\in\mathcal{B}$. But $$(U\cap U')\cap(V\cap V\,')=(U\cap V)\cap(U'\cap V\,')=B\cap B\,'\;,$$ so $B\cap B\,'\in\mathcal{B}$.

And this does show that $\mathcal{B}$ is a filterbase. (Indeed, it’s a rather nice one, since it’s actually closed under taking finite intersections.)
In addition to any changes needed to match the adjustments so far, your conclusion could be stated a bit more clearly:

Now let $U$ be any open nbhd of $a$. Clearly $X$ is an open neighborhood of $b$, so $U=U\cap X\in\mathcal{B}$. Thus, each open nbhd of $a$ contains (in fact is) an element of $\mathscr{B}$, and therefore $\mathcal{B}$ converges to $a$. Similarly, if $V$ is any open nbhd of $b$, then $V=X\cap V\in\mathcal{B}$, since $X$ is an open nbhd of $a$, and $\mathcal{B}$ converges to $b$ as well.

