'Cofinality' and base cardinality of convergent net less than cardinality of index set, implies constant from some index onwards? I have seen the following statement:

Let $\{ x_i\}_{i\in I}\subseteq \mathbb{R} $ be a net converging to an element $x_0\in \mathbb{R}$, where $\vert I\vert>\aleph_0$, $I$ is totally ordered and every lower set of $I$ is countable. Then there exists $i_0\in I$ such that $x_i=x_0$ for all $i\geq i_0$, i.e., $x_i$ is constant from some point onwards.

The solution\proof of this statement revolves around the fact that there exists a countable local base of $x_0$ and that a countable union of countable sets is countable.
It seems to me that using this argument, I can claim the following more general statement:

Let $X$ be a Hausdorff space and $x_0\in X$ such that $x_0$ has a local base $\mathcal{B}_{x_0}$ of cardinality less than $\kappa>\aleph_0$. Then any convergent net $\{ x_i\}_{i\in I} \subseteq X $ satisfying that $x_i\to x_0$, $\vert I\vert=\kappa$, and every lower set of $I$ has at most $\alpha<\kappa$ elements, must satisfy that $x_i=x_0$ from some point onwards.

I think this is proven by noticing that
$$ \bigcup_{U\in \mathcal{B}_{x_0}}\big( U^c\cap \{x_i \}_{i\in I}\big) = \{ x_i\}_{i\in I}\setminus \{x_0\}, $$
and hence
$$ \big\vert \{ x_i\}_{i\in I}\setminus \{x_0\}\big\vert\leq \vert \mathcal{B}_{x_0}\vert \cdot \alpha \leq \max\{ \alpha, \vert\mathcal{B}_{x_0}\vert \} <\kappa.$$
I would appreciate if anyone can verify whether this argument is generally sound. Secondly, I think this statement can be rephrased using cofinality.
I think that $\{ x_i\} $ is not constant from some index onwards if and only $\{ j\in I: x_i\neq x_0\} $ is cofinal and that we can also say that every cofinal set in $I$ would have to be of cardinality greater than $\vert \mathcal{B}_{x_0}\vert$ in the previous case. So I was wondering whether the following seems true:

Let $X$ be a Hausdorff space and $x_0\in X$ such that $x_0$ has a local base $\mathcal{B}_{x_0}$ of cardinality less than $\kappa>\aleph_0$. Then any convergent net $x_i\to x_0$ satisfying $\kappa>\text{cf}(I)>\vert \mathcal{B}_{x_0}\vert$ and $\vert I\vert\geq \kappa$, must be constant from some index onwards.

 A: The proof is correct but the `noticing' needs a bit more explaining and will reveal more about the ordered set $I$, without assuming the order is linear.

*

*For every $i\in I$ the complement, $L_i$ say, of $\{j:j\ge i\}$ is a lower set and hence countable. This helps in showing that $U^c\cap\{x_i:i\in I\}$ is of cardinality at most $\alpha$ for every neighbourhood $U$.

*If $A$ is a countable subset (in the original case), or of cardinality less than $\kappa$ (in the second case) of $I$ then $\bigcup_{i\in A}L_i$ is countable or of cardinality less than $\kappa$, respectively. If you take $j$ in its complement then $j$ is an upper bound for $A$. So if you have a small neighbourhood base $\mathcal{B}$ and you have $i_U\in I$ for $U\in\mathcal{B}$ then you can take an $i$ above all $i_U$ and the net will be constant above that $i$.

Your general statement is false: let $I$ be the ordinal $\omega_\omega$. Then the cardinality of $I$ is $\aleph_\omega$ and its cofinality is $\aleph_0$.
Define $x_i=2^{-n}$ if $\omega_{n-1}\le i<\omega_n$ (where $\omega_{-1}=-1$ for the nonce). This net converges to $0$ in $\mathbb{R}$ but it is not constant on a tail.
Part 2 The above reflects an earlier version of the question. The new final statement is correct but there is no need to introduce a cardinal $\kappa$. If $\mathrm{cf}(I)>|\mathcal{B}|$ then every net that converges to the point in question is constant on a tail: given $i_U\in I$ for $U\in\mathcal{B}$ we know that $\{i_U:U\in\mathcal{B}\}$ is not cofinal, so, as the order is linear, the set has an upper bound $j$ and then $\{x_i:i\ge j\}\subseteq\bigcap\mathcal{B}=\{x_0\}$.
Also, in the first statement $|I|=\aleph_1$, and in the second one $\kappa$ is the cardinal successor of $\alpha$. This follows from the general statement that a linear order in which all initial segments have cardinailty at most $\lambda$ has cardinality at most $\lambda^+$.
