$|A \cup B| = \mathfrak{c}$, then $A$ or $B$ has cardinality $\mathfrak{c}$. $A$ and $B$ are sets. $|A \cup B| = \mathfrak{c}$, prove that $A$ or $B$ has cardinality $\mathfrak{c}$.
This is an exercise problem from my textbook. It's easy if I assume CH to be true. But how can I prove this without it? I don't know how to eliminate the case of $A$ and $B$ both has cardinality larger than that of natural numbers and strictly smaller than that of reals.

It has been suggested in the comments that you should use the theory of well-ordered sets or Zorn's lemma. Here, instead, is a simple proof using the axiom of choice directly. 
I assume that you are familiar with the Cantor-Bernstein theorem. Because of that, it suffices to prove that $|A|\ge\mathfrak c$ or $|B|\ge\mathfrak c.$ I also assume you know that $|\mathbb R\times\mathbb R|=|\mathbb R|=\mathfrak c$; thus we may assume that $A\cup B=\mathbb R\times\mathbb R.$
Case 1. If $A$ is disjoint from some horizontal line in the plane $\mathbb 
R\times\mathbb R,$ then $B$ contains a horizontal line, and so $|B|\ge\mathfrak c.$
Case 2. If $A\cap L\ne\emptyset$ for every horizontal line $L,$ then by the axiom of choice there is a set $S\subseteq A$ such that $|S\cap L|=1$ for every horizontal line $L.$ Clearly $|S|=\mathfrak c$ and so $|A|\ge\mathfrak c.$
P.S. This argument shows that, if $a\lt c$ and $b\lt c,$ then $a+b\lt c\cdot c.$ This is a special case of Kőnig's theorem: if $a_i\lt b_i$ for each $i,$ then $\sum_i a_i\lt\prod_i b_i.$
