No, there is no purely algebraic proof of FTA.
So, as someone already noted, FTA is a misnomer.
I think the following proof is one of the most algebraic ones, though it's not purely algebraic.
Assumptions
We assume the following facts.
(1) Every polynomial of odd degree in $\mathbb{R}[X]$ has a root in $\mathbb{R}$.
(2) Every polynomial of degree 2 in $\mathbb{C}[X]$ has a root in $\mathbb{C}$.
Note:
(1) can be proved by the intermediate value theorem.
(2) can be proved by the fact that every polynomial of degree 2 in $\mathbb{R}[X]$ has a root in $\mathbb{C}$.
Notation
We denote by $|G|$ the order of a finite group $G$.
Lemma
Let $K$ be a field.
Suppose every polynomial of odd degree in $K[X]$ has a root in K.
Let $L/K$ be a finite Galois extension.
Then the Galois group $G$ of $L/K$ is a 2-group.
Proof:
We can assume that $L \neq K$.
By the theorem of primitive element, there exists $\alpha$ such that $L = K(\alpha)$.
By the assumption, the degree of the minimal polynomial of $\alpha$ is even.
Hence $|G|$ is even.
Let $|G| = 2^r m$, where $m$ is odd.
Let $P$ be a Sylow 2-subgroup of $G$.
Let $M$ be the fixed subfield by $P$.
Since $(M : K) = m$ is odd, $m = 1$ by the similar reason as above.
QED
The fundamental theorem of algebra
The field of complex numbers $\mathbb{C}$ is algebraically closed.
Proof:
Let $f(X)$ in $\mathbb{R}[X]$ be non-constant.
It suffices to prove that $f(X)$ splits in $\mathbb{C}$.
Let $L/\mathbb{C}$ be a splitting field of $f(X)$.
Since $L/\mathbb{R}$ is a splitting field of $(X^2 + 1)f(X)$, $L/\mathbb{R}$ is Galois.
Let $G$ be the Galois group of $L/\mathbb{R}$.
Let $H$ be the Galois group of $L/\mathbb{C}$.
By the assumption (1) and the lemma, $G$ is a 2-group.
Hence $H$ is also a 2-group.
Suppose $|H| > 1$.
Since $H$ is solvable, $H$ has a nomal subgroup $N$ such that $(H : N) = 2$.
Let $F$ be the fixed subfield by $N$.
Since $(F : C) = 2$, this is a contradiction by the assumption (2).
Hence $H = 1$. It means $L = \mathbb{C}$.
QED
Nope; every element of a finite field is algebraic over $\mathbf{F}_p$, but $\mathbf{F}_p(x)$ contains a element transcendental over $\mathbf{F}_p$... and thus so does its algebraic closure.
However, every field of characteristic $p$ can be written as a union of finite fields and copies of $\mathbf{F}_p(x)$. The easiest way is to just throw in a new copy of $\mathbf{F}_p(x)$ for every transcendental element of your field, obtained by mapping $x$ to that element.
Also, if $K$ is any field at all, the algebraic closure of $K$ can be written as a union of finite extensions of $K$. (which won't be finite fields, of course, unless $K$ is finite itself)
Best Answer
In general, an algebraic closure of a field $K$ is denoted by $\overline{K}$. Typical examples arising in number theory are $K=\mathbb{Q}$, $K=\mathbb{F}_p(t)$, $K=\mathbb{Q}_p$. Usually one needs the axiom of choice in order to prove the existence of algebraic closures. There are (at least) two exceptions: For $K=\mathbb{R}$ we have $\overline{K}=\mathbb{C}$. Then, for every subfield $K \subseteq \mathbb{R}$, we may realize $\overline{K}$ as the subfield of $\mathbb{C}$ which consists of complex numbers which are algebraic over $K$. It also exists without AC. We may also replace $\mathbb{R}$ by a real closed field, one only has to adjoin $\sqrt{-1}$. For $K=\mathbb{F}_q$, a finite field, we have for every $n$ an extension $\mathbb{F}_{q^n}$ of degree $n$ and every divisibility relation $n|m$ induces a canonical $\mathbb{F}_q$-homomorphism $\mathbb{F}_{q^n} \to \mathbb{F}_{q^m}$. It follows that we may consider the colimit $\mathbb{F}_{q^{\infty}} := \varinjlim_{n} \mathbb{F}_{q^n}$ (often this is written as a union, which is not quite correct). This turns out to be an algebraic closure of $\mathbb{F}_q$.
Let me also share a quite nice construction of an algebraic closure: Consider the (infinite) tensor product $A$ of all the $K$-algebras $K[x]/(f)$, where $f \in K[x] \setminus \{0\}$. By linear algebra it is non-zero, hence has a maximal ideal $\mathfrak{m}$ (Zorn's Lemma!). Then $K' := A/\mathfrak{m}$ is a field extension of $K$, and by construction every $f \in K[x] \setminus \{0\}$ has a root in $K'$. It is a nontrivial result that this already is the algebraic closure; but even if we don't use this, we can just repeat this process $K \hookrightarrow K' \hookrightarrow K'' \hookrightarrow K''' \hookrightarrow \dotsc$ and observe that the colimit $\overline{K}$ is an algebraic closure of $K$. A similar reasoning can be obtained to show that every two algebraic closures are isomorphic (but not in a canonical way).