Universal first countability in ZF?

Let $X=\mathbb{N}\times\mathbb{N}\cup\{\infty\}$, topologized such that a nonempty set is open iff it contains $\infty$ and contains cofinitely many points of $\{n\}\times\mathbb{N}$ for each $n\in\mathbb{N}$. Then $X$ is not first-countable at $\infty$. Indeed, suppose $(U_n)$ is a local base at $\infty$. For each $n$, let $j_n$ be the least $j$ such that $(n,j)\in U_n$ (since $U_n$ is nonempty and open, it must contain $(n,j)$ for all but finitely many $j$). Now let $$U=\{\infty\}\cup\{(n,j)\in\mathbb{N}\times\mathbb{N}:j>j_n\}.$$ Then $U$ is an open neighborhood of $\infty$. However, $U$ does not contain any $U_n$, since for each $n$, $(n,j_n)\in U_n$ but $(n,j_n)\not\in U$.

This is not just a pathological example--similar examples come up very naturally when you glue together infinitely many spaces at a point, for instance, when considering non-locally finite CW complexes in algebraic topology. As a very concrete example, a wedge sum of countably infinitely many copies of $[0,1]$ can be proven (in ZF) to not be first-countable at the basepoint by essentially the same argument.

Topologize $\mathbb N$ so that a subset $U$ of $\mathbb N$ is open if either $1\notin U$ or else $\sum\{\frac1n:n\in\mathbb N\setminus U\}\lt\infty.$
I claim that there is no countable base for the neighborhoods of $1.$

Suppose $U_1,U_2,U_3,\dots$ is an infinite sequence of open sets containing $1.$ For each index $i,$ since $U_i$ is an infinite set, we can define $n_i$ as the least element of $U_i$ which exceeds $2^i.$ Then, since $\sum_{i=1}^\infty\frac1{n_i}\lt\sum_{i=1}^\infty2^{-i}\lt\infty,$ the set $U=\mathbb N\setminus\{n_1,n_2,n_3,\dots\}$ is an open set containing $1;$ and for each $i$ we have $U_i\not\subseteq U$ since $n_i\in U_i\setminus U.$