A general form of Egorov (e.g. https://www.ime.usp.br/~glaucio/mat6704/textos/GMTLecureNotes.pdf) states that:
Egorov Theorem : Let $\mu$ be an outer measure on the set $X$ and $(Y,d)$ a separable metric space. Consider $A\subseteq X$ with $\mu(A)<\infty$ and a sequence of measurable functions $f_n:X\to Y$ in the sense that $f_n^{-1}(U)$ is $\mu$-measurable in $X$ for every open $U$ in $Y$. If $\{f_n\}$ converges pointwise $\mu$-a.e. on $A$ to a $\mu$-measurable function $f:\mathrm{dom}(f)\subseteq X\to Y$, then $\{f_n\}$ converges almost uniformly to $f$ on $A$ in the sense that for any $\epsilon>0$, there is a $\mu$-measurable set $B$ such that $\mu(A-B)<\epsilon$ and $\{f_n\}$ converges uniformly to $f$ on $B$.
Here, a set $A\subseteq X$ is called $\mu$-measurable if it satisfies the Carathéodory condition: $\mu(T)=\mu(T\cap A) + \mu(T-A)$ for any $T\subseteq X$.
My question: is there a counterexample to Egorov Theorem without separability? That is, is there a sequence of $\mu$-a.e. convergent functions valued in a non-separable metric space $Y$ that does not converge almost uniformly?
The proof of Egorov Theorem goes the usual way by (I) defining $$C_{i,j}:= \bigcup_{n\geq j} \{x\in X: d(f_n(x),f(x))>1/i\},$$ (II) using $\mu$-a.e. convergence to show that for each $i$ the intersection $\bigcap_{j} C_{i,j}$ has zero measure, and then (III) using $\mu(A)<\infty$ and continuity of measure to choose $J(i)$ so large for each $i$ that the exceptional set $\bigcup_i C_{i,J(i)}$ has small measure.
Separability condition on $Y$ is used only in step (I) to show that the sets $C_{i,j}$ are $\mu$-measurable. Indeed, a separable metric space $(Y,d)$ has a countable basis for its metric topology so that the Borel $\sigma$-algebra $\mathcal{B}(Y\times Y)$ (generated by open sets in product topology on $Y\times Y$) coincides with the product $\sigma$-algebra $\mathcal{B}(Y)\otimes \mathcal{B}(Y)$. Then, the pre-image of open set $(1/i,\infty)$ in $\mathbb{R}$ under continuous map $(x,y)\mapsto d(x,y)$ is open and measurable wrt product $\sigma$-algebra $\mathcal{B}(Y)\otimes \mathcal{B}(Y)$, so that its pre-image $C_{i,j}$ under measurable map $x\mapsto (f_n(x),f(x))$ is measurable.
I have found a counterexample to this technique, if $(Y,d)$ is not separable. For this, consider $Y$ a set with cardinality strictly larger than continuum, and $d$ the discrete metric. Then it can be shown that the diagonal $\Delta \in \mathcal{B}(Y\times Y)- \mathcal{B}(Y)\otimes \mathcal{B}(Y)$; this is known as the Nedoma's paradox cf. https://www.drmaciver.com/2006/04/journal-of-obscure-results-1-nedomas-pathology/. Now put $X=Y\times Y$ and $f,g:Y\times Y\to Y$ projections onto the first and second coordinates respectively. By construction, $f,g$ are measurable, but the set $\Delta=\{(x,y)\in Y\times Y: d(f(x,y),g(x,y))<1\}$ is not measurable.
It seems more difficult to construct a counterexample to the Egorov statement... Any help is appreciated :)