X is Hausdorff iff a net converges to at most one point I wrote proof to following theorem. I ask for verification of my reasoning.
Theorem 3 from General Topology by Kelley:

A topological space is a Hausdorff space if and only if each net in the space converges to at most one point.

Proof
"$\Rightarrow$" Let $(X,\tau)$ be a Hausdorff space. For any points $x,y$ of topology $\tau$ there are $Z,W\in \tau\quad$  s.t.
$x\in Z $ and $y\in W$, and $Z\cap W \in \tau$. Hence there is a net $S_n$ in $X$ directed by $\subset$.
Since $X$ is a Hausdorff space, there are neighborhoods $U,V$ in base of the space s.t
$x\in U \subset Z$,
$y\in V\subset W$, and $U\cap V = \emptyset$.
So if $S_n$ converges to both $x$ and $y$, it is eventually in $U$ and $V$, which are disjoint. Therefore $S_n$ can converge to one point at most.
"$\Leftarrow$" Let each net $S_n$ converge to at most one point in the space $X$.
Suppose that $S_n$ converges to both $x$ and $y$.
Then $S_n$ is eventually in both neighborhoods of $x$ and $y$. Hence there are neighborhoods of two points which are not disjoint. So $X$ is not a Hausdorff space.
Is following conclusion valid, to complete the proof?
Therefore if $S_n$ converges to at most one point in $X$, it is a Hausdorff space.
 A: It seems to me you have the right ideas but your proof needs a bit of cleaning.
Here is a sketch that may help you clean your proof:

for any $x\in X$, denote by $\mathcal{V}_x$ the collection of open sets containing $x$.
Suppose $X$ is a topological space where each convergent net has a unique limit. If $X$ were not Hausdorff, then there would exist a pair of points $x$ and $y$ such that for any open sets $V\in\mathcal{V}_x$ and $U\in\mathcal{V}_y$  there is $x_{U,V}\in V\cap U$.  Then  $\{x_{V,U}:(V,U)\in\mathcal{V}_x\times\mathcal{V}_y\}$ is a net in $X$ (why?) that converges to both $x$ and $y$ which is a contradiction.
Conversely, suppose $X$ is Hausdorff and $\{x_n:n\in D\}$ is net converging to $x$ and $y$. If $x\neq y$, let $V_x$ and $V_y$ be disjoint open neighborhoods of $x$ and $y$ respectively. There is $m\in D$ such that $x_n\in V_x$ and $x_n\in V_y$ for all $n\geq m$. This is a contradiction to $V_x\cap V_y=\emptyset$.
A: Your proof is substantially correct: to follow (I hope you like it) I rewrite it with some additional words so that it could must more clear.
Theorem
A topological space $X$ is hausdorff if and only if any net $(x_\lambda)_{\lambda\in\Lambda}$ converges at most one point $x$.
Proof. So let be $X$ is hausdorff and we suppose that $(x_\lambda)_{\lambda\in\Lambda}$ is a net converging to the distinct point $x$ and $y$. So if $X$ is hausdroff and if $x$ and $y$ are different point of $X$ then there exist two open and disjoint sets $U_x$ and $U_y$ containing respectively $x$ and $y$ so that if $(x_\lambda)_{\lambda\in\Lambda}$  converges to $x$ and $y$ then tere exist $\lambda_x,\lambda_y\in\Lambda$ such that $x_\lambda\in\ U_x$ for any $\lambda\ge\lambda_x$ and $x_\lambda\in U_y$ for any $\lambda\ge\lambda_y$ and so for $\lambda_{xy}\in\Lambda$ such that $\lambda_{xy}\ge\lambda_x,\lambda_y$ (remember that $\Lambda$ is a directed set) it follows that $x_\lambda\in U_x\cap U_y$ for any $\lambda\ge\lambda_{xy}$ and clearly by the definition of $U_x$ and $U_y$ this is impossible.
Now we suppose that there exist a net $(x_\lambda)_{\lambda\in\Lambda}$ in $X$ converging to two different points $x$ and $y$ so that through the same argument promoted above it folows that there exist a $\lambda_0\in\Lambda$ such that $x_\lambda\in U_x\cap U_y$ for any open neighborhood $U_x$ and $U_y$ of $x$ and $y$ and so clearly $X$ is not hausdorff. So we conclude that if any net $(x_\lambda)_{\lambda\in\Lambda}$ in $X$ converges at most one point then $X$ is hausdorff.
