One should note that this argument fails for uncountably many wizards. A strategy for $2^{|\mathbb N|}$ wizards that succeeds with probability $100\%$ (assuming this probability is well defined) is presented below. The strategy given here by xnor can be extended fairly naturally. To summarize the strategy there for $2^{n}-1$ wizards:
The wizards are put into bijection with the non-empty subsets of $\{1,2,\ldots,n\}$. Then, we compute a new set $S$ by letting $s\in S$ if, there are an odd number of wizards who have black hats and whose associated sets include $s$. Any wizard who can guess their own hat color in a way that would make $S$ empty will make the opposite guess.
And the lemmas one might need to show that this strategy is effective would be noting that $S$ can be any subset of $\{1,\ldots,n\}$ with equal probability and that, if it is non-empty, one wizard will make a guess (in particular, the wizard associated with the set $S$).
To extend this, let $W$ be the set of wizards. Choose some bijection $f:W\rightarrow P(\mathbb N)\setminus\{\emptyset\}$. Next, let $E$ be some function $E:P(W)\rightarrow \{0,1\}$ such that, if two sets $S$ and $S'$ differ by one element (i.e. $|(S\setminus S')\cup (S'\setminus S)|=1$) then $E(S)=1-E(S')$. The existence of such an $E$ follows from the axiom of choice (in the finite case, we can just use $E(S)=|S|\text{ mod }2$, which breaks down for infinite $S$). Then, consider the set $S\subseteq \mathbb N$ defined as
$$S=\{n\in\mathbb N : E(W_n)=1\}$$
where $W_n$ is the set of wizards $w$ whose hats are black and for which $n\in f(w)$. Each wizard guesses that $S$ is not empty and thus, if one of their choices would make $S$ zero, they do not make the guess. Notice that, if $S$ is not empty, then changing the hat color of the wizard associated with $S$ would make it empty - thus, that wizard is able to deduce the color of their hat if they know $S$ is not empty.
This strategy therefore lets a single wizard guess correctly whenever $S$ isn't empty. Noting that there is a unique arrangement of the hat colors on the wizards associated with the sets $\{1\},\,\{2\},\,\{3\},\ldots$ that makes $S$ empty given the hat colors of everyone else, and this arrangement must be achieved with probability $0$ as it requires infinitely many independent events of probability $\frac{1}2$ to occur. Partitioning the wizards into countably many uncountable groups and applying this strategy gives countably many correct guesses with probability $1$.
(I note "assuming the probability is well defined" because the mathematical formalisms handling probability tend to not play well with those handling the axiom of choice; in particular, it's not obvious that the set of positions for which this strategy wins is measurable, which would mean no probability could be assigned to it. It's possible that such strategies exist for countably many wizards too)
