In Field theory by Steven Roman Chapter 9 Exercise 20, if we write the algebraic closure of finite field $F_q$ as $\Gamma(q)$ and $a_n$ be any strictly increasing infinite sequence of positive integers, the exercise wants to prove that $\Gamma(q)=\bigcup_{n=0}^{\infty}GF(q^{a_n})$.
However, if $a_n$ is an arbitrary sequence, we are even unable to prove $\bigcup_{n=0}^{\infty}GF(q^{a_n})$ is a field. I wonder whether the exercise has omitted some condition since the equality doesn't hold under current conditions offered.
Hope for answers!
Since the question is now closed on MathOF (and rather belongs on here), here's an answer in the form of an exercise ((a),(b) below). Yes, the exercise in Roman's book is false and it is quite hardly reassuring that such an exercise appears as published in a GTM book.
First, to be rigorous, we should work within one given algebraically closed field $K$ with characteristic $p$ (where $p$ is the unique prime divisor of $q$), so that $\mathrm{GF}(q^k)$ is a well-defined subfield of $K$, and so that the union makes sense as a subset of $K$.
(a) As suggested by @reuns (and copied by the OP to the MO post without reference), for a sequence $a=(a_n)$, the union $L_a=\bigcup_n\mathrm{GF}(q^{a_n})$ is an algebraic closed subfield of $K$ iff for every $k\ge 1$ there exists $n$ such that $k$ divides $a_n$;
(b) this union $L_a$ is a subfield of $K$ iff for any $m,m'$, there exists $n$ such that $\mathrm{lcm}(a_m,a_{m'})$ divides $a_n$.
(c) for every subfield $L$ of $K$ algebraic over $\mathrm{GF}(q)$, there exists a sequence $a=(a_n)$, such that $a_n$ divides $a_{n+1}$ for every $n$, such that $L=L_a$.
For instance, if $a_n=2^n$ for all $n$, then the union is a field, but not algebraically closed. If $a_n$ is the $n$-th prime, the union is not a field.