I'm using the notation from this question.
For all sets $A,B$, $|A\cup B|\le_c |A|+|B|$, and if $A\cap B=\emptyset$ then $|A\cup B|=_c |A|+|B|$.
My attempt to prove it:
By definition, $$|A|+|B|=||A|\coprod|B||.$$ $|A\cup B|\le_c |A|+|B|$ means that there is an injection from $|A\cup B|$ to $|A|+|B|=||A|\coprod|B||$. Since for any set $S$, $S=_c |S|$, it suffices to show that there is an injection from $A\cup B$ to $|A|\coprod |B|=(\text{blue}\times|A|)\cup (\text{red}\times|B|)$. (Here $\text{blue}=\{blue\}$ and similarly for $\text{red}$.)
For a set $S$, let $f_S: S\to |S|$ be a bijection. Define a function $$f:A\cup B\to |A|\coprod |B|$$ as follows. If $x\in A$, then $f(x)=(blue,f_A(x))$. If $x\in B$, then $f(x)=(red,f_B(x))$. It is obviously injective as long as it is well defined. But why is it well-defined? If $x\in A\cap B$, then $f(x)=(blue,f_A(x))$ and $f(x)=(red,f_B(x))$...
If $A\cap B=\emptyset$, then $f$ is well-defined and injective. How to show it is surjective (if it is)?