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 $|A| = |B|$? 
 A: 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.
A: Let $U=\bigcup\limits_{A\in\mathcal A}A$. If one knows a bijection $A:\mathbb N\to\mathcal A$, hence $\mathcal A=\{A(n)\mid n\in\mathbb N\}$, and if, for every $A(n)$ in $\mathcal A$, one knows a bijection $B_n:\mathbb N\to A(n)$, then the function $C:\mathbb N\times\mathbb N\to U$, $(n,k)\mapsto B_n(k)$, is a surjection. If $\mathcal A$ is made of disjoint sets, $C$ is a bijection. 
In the general case, $|U|\leqslant|\mathbb N\times\mathbb N|=|\mathbb N|$. Since $A(0)\subseteq U$ and $|A(0)|=|\mathbb N|$, this is enough to deduce that $|U|=|\mathbb N|$. 
When $\mathcal A$ is not made of disjoint sets, one could get an explicit bijection between $\mathbb N\times\mathbb N$ and $U$ based on the bijections $B_n$, using $B_n$ only on the elements of $A(n)$ not already in $\bigcup\limits_{k\lt n}A(k)$, and concatenating recursively the result, but to write down completely this construction would be somewhat messy.
A: Q1. Yes, see Cantor-Bernstein's theorem. 
Q2. Yes, this is the definition of $|A|=|B|$. And, $|A|\le |B|$ is defined as there is an injection $A\to B$, and the Cantor Bernstein theorem states that $|A|\le |B|\le |A| \implies |A|=|B|$ (using the axiom of choice).
Anyway, in your case the same method works as for the denumerablility of $\Bbb Q$, or $\Bbb N\times\Bbb N$, and no need for the general Cantor-Bernstien.
A: Let $(A_i)_{i \ge 0}$ be a countable family of sets and $A = \cup A_i$.
For each $i \ge 0$ let there be given $([a_i]_k)_{\,k \ge 1}$, a bijective enumeration of the set $A_i$.
We will be using the following result (see it also used over here).
Proposition 1: A set $A$ is countably infinite if there exist a family of subsets of $A$,
$(A_n)_{\, n \ge 0}$ satisfying
$\quad |A_n| = n$
$\quad A_k \subset A_{k+1} \; \text{ for } k \ge 0$
$\quad A = \bigcup_k A_k$
Moreover, the chain $A_0 \subset A_1 \subset A_2 \dots$ defines an explicit bijective enumeration of the set $A$.
Proof
Given such a chain of finite subsets, for every $k \ge 1$ there exist one and only one element $a_k \in A_k$ such that $a_k \notin A_{k-1}$. One can also easily show that $(a_k)_{\,k \ge 1}$ is a bijective enumeration of $A$.
$\quad \blacksquare$
For each $i \ge 0$ and $m \ge 1$ we define
$\quad A_{(i,m)} = \{ x \in A_i \,| \, x \in \{([a_i]_1,([a_i]_2,\dots,([a_i]_m\}\}$
We define
$$\quad R_m = \bigcup_{i=1}^m A_{(i,m)}$$
If $B$ is any subset of $A$ and if $R_m$ is not included in $B$ we can apply an operator $\mathcal R_m$ to $B$ as follows:
$\quad B \mapsto B \cup \{\hat r\}$
where $\hat r$ is 'cranked-out' as follows:
$\text{1. }$ Find the smallest integer $k$ with $1 \le k \le m$ such that $A_{(k,m)}$ is not contained in $B$.
$\text{2. }$ Find the smallest integer $j$ with $1 \le m \le m$ such that $[a_k]_j \notin R_m$; set $\hat r = [a_k]_j$.
If $C$ is any proper subset of $A$ we define $\mathcal M$ on $C$ as follows:
$\quad C \mapsto m \; \text{ where } m \text{ is the smallest positive integer such that } R_m \text{ is not included in } C$
All the constructions are now available to define the recursion. With $D_0 = \emptyset$ and given $D_k \subsetneq A$ with $k \ge 0$, define 
$$   D_{k+1} = \mathcal R_{\mathcal M (D_k)}(D_k)$$
It is easy to see that this chain satisfies the conditions stated in proposition 1.
A: 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.
