Explicit bijection between equipotent sets? I'm thinking about the proof of the following theorem:

If $\mathcal A$ is a denumerable family of denumerable sets then $\bigcup \mathcal A$ is denumerable. (denumerable means that there is a bijection to $\mathbb N$)

The proof shows $|\mathbb N| \leq |\bigcup \mathcal A|$ and $|\mathbb N| \geq |\bigcup \mathcal A|$ rather than giving an explicit bijection $f: \mathbb N \to \bigcup \mathcal A$. 

Question 1: In this case, is it possible to give an explicit bijection?

Question 2: In general, is it possible to find a bijection between set $A$ and $B$ if we know that $|A| = |B|$?
First note that the Cantor-Bernstein is constructive and does give us a bijection.

Secondly, if we require the sets in $\cal A$ are pairwise disjoint then we can explicitly define a map by sending $A_i$ to $\{i\}\times\Bbb N$ and composing the whole thing with the Cantor pairing function. However if we don't have this assumption then we cannot do it that way because the function from the union is not well-defined anymore. We could require that an element is mapped to $\langle i,k\rangle$ where $i$ is the least index such that $a\in A_i$, but then the function is not a bijection anymore.

For this reason it is often simpler just to show mutual bijections (or in the $\mathbb N$ case, a surjection from $\mathbb N$ onto the union is also enough).

As for your second question, the answer is no. By the assumption that $|A|=|B|$ we can prove that there exists a bijection between $A$ and $B$, but it is not necessary that we can define it. This of course, depends on the meaning of "define", but if we take it to mean write an explicit formula that the collection of sets satisfying it is a bijection between $A$ and $B$, then the answer is no.

For example, if we can write down a definitive bijection between $\mathbb R$ and some ordinal then we invariably solve the continuum hypothesis. We could of course parameterize, but that would depend on parameters which are not definable themselves. In fact, if your underlying theory is merely ZF then such bijection would also prove that $\mathbb R$ can be well-ordered, which cannot be done without choice.
The answer to question one is given in did's answer and is yes, in this case we can write down a bijection explicitly, even if we don't assume $A$ and $B$ to be disjoint.

The answer to question two is given in Asaf's answer and is no, in general we cannot expect to write down the bijection even if we know that it exists. For example, let's assume we know that there is a bijection $f: \mathbb R \to \omega_2$. Since $\omega_2$ is an ordinal it is well-ordered so that $\mathbb R$ would automatically also be well-ordered, without the need for AC. But we know that $\mathbb R$ is not well-ordered unless we assume AC.