If $\omega \le m^2$ then $\omega \le m$ without AC?

Solution 1:

Given an infinite sequence $((a_i,b_i))_{i \ge 0}$ of distinct elements of $X \times X$ (where $X$ is a set of cardinality $m$), form the sequence $(a_0,b_0,a_1,b_1,a_2,b_2,...)$. This sequence is an infinite sequence, which may, however, contain some repetitions. Let $Y$ be the set formed by the terms of this sequence (without repetitions). Then, $Y$ must be infinite, because otherwise, the sequence $((a_i,b_i))_{i \ge 0}$ would contain infinitely many distinct elements of the finite set $Y \times Y$, which is a contradiction. So, removing duplicates still gives an infinite sequence, of distinct elements of $X$ this time.