The Brownian zero set is uncountable

Fix a rational $q>0$ with $q\mathrm{\notin }Z$; then $Z\cap [0,q]$ is nonempty because $0\in Z$, it is closed in $\mathbb{R}$ as the intersection of the closed set $Z$ with the closed interval $[0,q]$, and it has no isolated points: for $t\in Z\cap [0,q]$ one has $t<q$, and by [F2] there are zeros of $Z$ different from $t$ in every neighbourhood of $t$, which for a neighbourhood of radius $<\min (q-t,t+1)$ lie in $[0,q]$.

Almost surely, for every $T>0$ there is a rational $q\in (0,T)$ with $q\mathrm{\notin }Z$: by [F3] and monotonicity of Lebesgue measure, $\lambda (Z_T)=0<T=\lambda ([0,T])$, so $Z_T\mathrm{\ne }[0,T]$; picking $s\in (0,T)\setminus Z$, openness of the complement of the closed set $Z$ supplies a neighbourhood of $s$ disjoint from $Z$, and [F5] supplies a rational $q$ in that neighbourhood with $0<q<T$.

On the probability-one event of [step 1.2] and [F2], and for a rational $q$ as there, [step 1.1] exhibits $Z\cap [0,q]$ as a nonempty perfect subset of $\mathbb{R}$; by [F4] it is uncountable, and since $Z\cap [0,q]\subseteq Z_T$, the set $Z_T$ is uncountable.

The cases are covered: $T>0$ is required, so the interval is nondegenerate; the rational $q$ is chosen strictly inside $(0,T)$ so that the potential isolated point $q$ of $Z\cap [0,q]$ is excluded by $q\mathrm{\notin }Z$; the statement is asserted simultaneously for all $T>0$ on one probability-one event, obtained by intersecting the countably many events of [step 1.2] over rational $T$ and using monotonicity in $T$; and AC enters only through [F6].