A net converges to a point iff every subnet accumulates in that point.

convergence-divergencegeneral-topologynetssolution-verification

While working on a takehome for my functional analysis course I stumbled upon this small lemma

A net $$(x_i)_{i\in I}$$ in a topological space $$X$$ converges to a point $$x\in X$$ if and only if every subnet has a accumulation point in $$x$$.

This is a slightly stronger formulation of the following well known result in topology.

A net $$(x_i)_{i\in I}$$ in a topological space $$X$$ converges to a point $$x\in X$$ if and only if every subnet converges to $$x$$.

I managed to come up with the following proof, but I doubt my judgement because it seems a little unbelievable for me to come up with a stronger version of an existing mathematical result. Can you check my proof?

the implication from left to right is trivial because if $$(x_i)_{i\in I}$$ converges to $$x$$ then so will any subnet.
Convergence to $$x$$ implies that the subnet has a accumulation point in $$x$$ as well, because this is a weaker statement.

Now if $$(x_i)_{i \in I}$$ does not converge to $$x$$, it has a subnet which does not converge to $$x$$, $$(x_{\sigma(j)})_{j\in J}$$.
This means there is an open neighbourhood $$U$$ of $$x$$ such that for any $$j \in J$$ there exists a $$j' \geq j$$ such that $$x_{\sigma(j')} \not\in U$$.
Using the map
$$J \to J : j \to j'$$
we find the subnet $$(x_{\sigma(j')})_{j \in J}$$ which has no accumulation point in $$x$$.

The net $$(x_j)_j$$ has $$x$$ as an accumulation point iff it has a subnet that converges to $$x$$.