Let $X$ be a non-empty set and $I=[1, \alpha]$ for some $\alpha>1$.
Suppose that $(X, \tau_i)_{i\in I}$ is a family of topological spaces such that $\tau_i\supseteq\tau_j$ whenever $i\leq j$.
Question: Is there a topology $\tau$ on $X$ such that, for any net $(x_j)$ in $X$ with limit $x\in X$, $$\tag{1} (x_j)\rightarrow_{\tau} x \quad\text{iff:}\quad (x_j)\rightarrow_{\tau_j} x \ \text{ for } \textbf{ some } j\in [1, \alpha) \ ?$$
(The candidates $\tau:=\tau_\alpha$ and $\tau:=\cap_{j\in J}\tau_j$ don't give the $(\Rightarrow)$-direction in (1).)
No. If $\tau$ and $\tau'$ are two topologies on a set, then $(x_j)\to_\tau x$ implies $(x_j)\to_{\tau'}x$ for all nets $(x_j)$ and all points $x$ in $X$ iff $\tau'\subseteq\tau$. So, the reverse implication of (1) means that $\tau$ must be contained in $\bigcap_{j\in[1,\alpha)}\tau_j$. Thus if you have any example where $\tau=\bigcap_{j\in[1,\alpha)}\tau_j$ does not satisfy the forward direction of (1), neither can any coarser topology, and so no topology can satisfy both directions of (1).
Explicitly, for instance, let $X=[1,2]$ and let $\tau_j$ be the topology where a set $U$ is open iff either $2\not\in U$ or $[1,j)\subseteq U$. Then $\bigcap_{j\in[1,2)}\tau_j=\tau_2$ has the property that the sequence $(2-1/n)_{n\in\mathbb{Z}_+}$ converges to $2$ but no $\tau_j$ for $j<2$ has this property.