Hausdorff space if no net converges to two different values Having trouble to prove the following statements regarding topological spaces:

Let $(X,\mathcal T)$ be a topological space.
  a) Show that for all mutually distinct points $x,y \in X$ there are open sets $U,V$ with $U\cap V = \emptyset$ and $x\in U, y \in V$ if, and only if, no net in $X$ converges to two different values.
  b) Let $\mathcal T'$ be a topology on $X$ and for all nets in $X$ it holds that they converge in $(X, \mathcal T)$ if, and only if, they converge in $(X, \mathcal T')$. Show that $\mathcal T = \mathcal T'$.

Anyone can help?
 A: Suppose that all distinct $x$ and $y$ have disjoint neighbourhoods, and suppose that $(x_i)_{i \in I}$ is a net converging to both $x$ and $y$. By assumption on $X$ find open $U$ with $x \in U$ and open $V$ with $y \in V$ such that $U \cap V = \emptyset$.
Now apply the definition of convergence to $x$ and $U$. 
So there exists $i_0 \in I$ such that ... 
Same for $y$ and $V$, getting a $j_0 \in I$ such that ... 
Now (as $I$ is directed!) find $k \in I$ with $k \ge i_0, k \ge j_0$. Where must $x_k$ be?
For the other direction, suppose every convergent net in $X$ has a unique limit.
Suppose now that the disjoint neighbourhood property fails for $X$, so there are $x \neq y$ in $X$ such that for all neighbourhoods $U$ of $x$ and all neighbourhoods $V$ of $y$ we have $U \cap V \neq \emptyset$.
Now define a net: $I = \{(U,V): U \in \mathcal{N}_x, V \in \mathcal{N}_y \}$, ordered by reverse inclusion in both coordinates, and let $f((U,V))$ be any point in $U \cap V \neq \emptyset$ (yes, we use the axiom of choice!). This $f$ defines a net in $X$ (i.e. a map from $I$, the directed set, to $X$; $f(i)$ is denoted $x_i$ as well if the context is clear, as I did above).
Now check (!) that this net converges to both $x$ and $y$ and we have our contradiction.
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If two topologies $\mathcal{T}_1,\mathcal{T}_2$ on $X$ are different, there is at least one subset that is open in one, say $O \in \mathcal{T}_1$, but not in the other, so $O \notin \mathcal{T}_2$. So there is some $x \in O$ that is not an interior point w.r.t. $\mathcal{T_2}$. This means that for every neighbourhood $N$ of $x$ (in $\mathcal{T}_2$), we have $N \nsubseteq O$, so there are always points in $N \setminus O$.
Now there is a standard filter that can now be constructed that converges to $x$ in $\mathcal{T}_2$ and does not converge to $x$ under $\mathcal{T}_1$...    
A: It is a known result that $(X, \mathcal{T})$ is Hausdorff if and only if the diagonal $\Delta = \{(x,x) : x \in X\}$ is a closed set in $X \times X$ with respect to the product topology.
Let $((x_\lambda, x_\lambda))_{\lambda\in\Lambda}$ be a net in $\Delta$ which converges to $(x,y) \in X \times X$. Then, by the property of the product topology, $(x_\lambda)_{\lambda \in \Lambda}$ converges to both $x$ and $y$ in $X$.
If $X$ is Hausdorff, then $\Delta$ is closed, so $(x,y) \in \Delta$ which implies $x = y$, therefore the limit is unique. Conversely, if nets in $X$ have unique limits, then necessarily $x = y$ so $(x,y) \in \Delta$, implying that $\Delta$ is closed in $X \times X$.

Assume that $x_\lambda \xrightarrow{\mathcal{T}_1} x$ implies $x_\lambda \xrightarrow{\mathcal{T}_2} x$.
We have:
\begin{align}
A \text{ is closed in } \mathcal{T}_2 &\iff A \text{ contains all limits of its convergent nets with respect to }\mathcal{T}_2\\
&\implies A \text{ contains all limits of its convergent nets with respect to }\mathcal{T}_1\\
&\iff A \text{ is closed in } \mathcal{T}_1
\end{align} 
Therefore, all closed sets in $\mathcal{T}_2$ are also closed in $\mathcal{T}_1$. By taking complements, we conclude $\mathcal{T}_2 \subseteq \mathcal{T}_1$.
