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.
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 $V_x\cap V_y=\emptyset$.