Algebraic Closure of Finite Field $\mathbb{F}_p$ Dummit and Foote (in Chapter 14.3) construct the algebraic closure of the finite field $\mathbb{F}_p$ by the following union: $$\bar{\mathbb{F}}_p = \bigcup_{n \geq 1}\mathbb{\mathbb{F}}_{p^n}.$$I'm having trouble seeing how we can take this infinite union, because to do so we have to view all the fields $\mathbb{F}_{p^n}$ as subsets of some larger object. Finite unions of the form $$\bigcup_{k=1}^n \mathbb{F}_{p^k}$$ make sense because all the fields in this union can be seen as subfields of a finite field of order $p^{n!}$, but I'm having trouble with the infinite case.
The authors write:

...given any two finite fields, $\mathbb{F}_{p^{n_1}}$ and $\mathbb{F}_{p^{n_2}}$ there is a third finite field containing (an isomorphic copy of) them, namely $\mathbb{F}_{p^{n_1n_2}}$. This gives us a partial ordering on these fields and allows us to think of their union.

How does the partial ordering give us a well defined union in the infinite case? Does it involve something like Choice or the principle of recursive definition? Is it true in general that a partial ordering on a set gives us a well-defined notion of a union even when the elements aren't literally sets contained in a larger set?
 A: As explained in the comments, the correct setup here is that of direct  limit (or filtered colimit). 
Given a preordered set $(I,\leq)$ and a direct system $(A_i)_i$ (so, with maps $A_i\to A_j$ for $i\leq j$) of algebraic gadgets (this is more general, but for algebraic things this always works), you can create their direct limit. 
If all the $A_i\to A_j$ are inclusions, you can think of this direct limit as a union (because each $A_i\to \varinjlim_i A_i$ will be injective, and any element of the direct limit comes from at least one $A_i$; and because if you then regard $A_i$ as a subset of $\varinjlim_i A_i$, with this identification, then the limit is literally the union)
So here, what is meant, is to take the divisibility order on $\mathbb N$ : $n\preceq m$ if and only if $n\mid m$. Then you have to chose an embedding $\mathbb F_{p^n}\to \mathbb F_{p^m}$, and you have to make these choices sort of consistently (there are ways to ensure this here, e.g. using factorials)
Then this forms a direct system, and its direct limit is the algebraic closure. And then if you identify each $\mathbb F_{p^n}$ with its image in the direct limit, then it's just their union. 
