Let $X$ be a set and $\mathcal{T_1,T_2}$ two Hausdorff topologies on $X$ such that they admit the same convergent nets, i.e., a net $(x_{\alpha})_{\alpha}$ in $X$ converges with respect to $\mathcal{T_1}$ iff it converges with respect to $\mathcal{T_2}$ (but to a possibly different point). Does it follow that $\mathcal{T_1=T_2}$?
If we require that convergent nets converge to the same point, then the result follows easily. If we don't require Hausdorff-ness, then the result is simply false, as the Sierpiński topology and the trivial topology on the set $\{a,b\}$ shows (where every net converges to some point). However, I wonder whether it's true for Hausdorff spaces?
Let $(x_{\alpha})_{\alpha\in A}$ be a net converging in to $x$ in $\mathcal{T}_{1}$ and to $y$ in $\mathcal{T}_{2}$. We define $$\hat{A}=A\times\{1,2\}$$ and we say $(\alpha,i)\leq(\beta,j)$ iff $\alpha\lneq\beta$ or $\alpha=\beta$ and $i\leq j$. Note that $\hat{A}$, $A\times\{1\}$ and $A\times\{2\}$ are directed sets. We define the net $(y_{(\alpha,i)})_{\hat{A}}$ by $$y_{(\alpha,i)}=\begin{cases}x_{\alpha}&\text{ if }i=1\\ y&\text{ if }i=2\end{cases}.$$
Clearly $(y_{(\alpha,i)})_{\hat{A}}$ converges to $y$ in $\mathcal{T}_{2}$, but in $\mathcal{T}_{1}$ the subnet $(y_{(\alpha,i)})_{A\times\{1\}}$ converges to $x$ and the subnet $(y_{(\alpha,i)})_{A\times\{2\}}$ converges to $y$.
Thus a net converges in $\mathcal{T}_{1}$ if and only if it converges in $\mathcal{T}_{2}$ convergent nets converge to the same point.