Jech notes in Chapter 5 of Set Theory, 2003 that, given a family of nonempty sets $S$, in some cases the existence of choice function can be proved in ZF:
(iii) when every X ∈ S is a finite set of real numbers; let f (X) = the least element of X.
However, in the very next sentence it is stated:
... one cannot prove existence of a choice function (in ZF) just from the assumption that the sets in $S$ are finite; even when every $X \in S$ has just two elements (e.g., sets of reals), we cannot necessarily prove that $S$ has a choice function.
(emphasis mine)
How is this not a contradiction with (iii)?
The real numbers are linearly ordered. So given a [non-empty] finite set, it has a well-defined minimum and maximum.
This allows us to define a uniform choice from all the [non-empty] finite subsets of the real line: simply pick the minimum of each set.
However, not every set is a set of real numbers. And not every finite set is a finite set of real numbers. And indeed it is consistent that there is a countable family of sets of size $2$ without a choice function. As a consequence, it means that if $S$ is a set such that all of these pairs are subsets of $S$, then $S$ cannot be linearly ordered.
The boldface part is mentioning that the elements of the finite sets are themselves sets of reals. Not real numbers, and not finite set of real numbers. And so, there is no longer a canonical way of choosing an element from each finite set.
Therefore, it follows, also, that with only $\sf ZF$, not every set can be linearly ordered.