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

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 statements are trivially equivalent if you know the standard fact:

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

So the second version of the statement (which you claim to know) is immediately equivalent to your lemma.