Re: (4). As you are only "95% sure".....In ZF, without analysis, we have
(i). If $X$ is a compact space and $f:X\to [0,\infty)$ is continuous and $y=\inf_{x\in X}f(x)$ then $\exists x\in X\;(f(x)=y).$ Otherwise $\{f^{-1}(y+r,\infty):r>0\}$ is an open cover of $X$ with no finite sub-cover.
(ii).(Edited). $[-r,r]^2$ is compact for positive real $r$. We derive this from (iii) below. [I don't want to re-number all the references in the second half of this A.]
(iii). A closed bounded real interval is compact.
(iv). (Elementary). For $r>0$ the set $R(r)=\{z\in \mathbb C: \max (|Re (z)|, |Im(z)|)\leq r\}$ is homeomorphic to $[-r,r]^2.$ By (ii) and (iii), $R(r)$ is compact.
(v). (Elementary). If $p:\mathbb C\to \mathbb C$ is a non-constant polynomial then $|p(z)|\to \infty$ uniformly as $|z|\to \infty$. And by elementary algebraic means, $|p|$ is a continuous function.
Theorem. A non-constant polynomial $p:\mathbb C\to \mathbb C$ has a zero. PROOF: The case deg $(p)=1$ is trivial, so assume deg$(p)>1.$
By (v) there exists $r>0$ such that $\forall z \in \mathbb C$ \ $R(r)\;( |p(z)|>|p(0)|).$
Let $R(r)$ be as in (iv). Now $R(r)$ is compact so there exists $z_0\in R(r)$ with $|p(z_0)|=\min \{|p(z)|:z\in R(r)\}.$
We have $|p(z_0)|\leq |p(0)|.$ Also $z\not \in R(r)\implies |p(z)|>p(0)|\geq |p(z_0)|.$ Therefore $$|p(z_0)|=\min \{|p(z)|:z\in \mathbb C\}.$$
We now assume $p(z_0)\ne 0$ and obtain a contradiction.
Since deg$(p)>1$ there exists (unique) $k\in \mathbb N$ and $A\ne 0$ and polynomial $q$ such that $$\forall z\;(p(z)=p(z_0)+A(z-z_0)^k(1+(z-z_0)q(z-z_0)).$$ There exists $y$ such that $Ay^k/p(z_0)$ is a negative real number. For such $y,$ we have $A(ys)^k/p(z_0)<0$ for all $s>0.$ Now there exists (small ) $s>0$ such that $$A(ys)^k=-tP(z_0)\text { for some } t\in(0,1)$$ $$\text { and }\quad |ys\cdot q(ys)|<1/2$$ . $$\text {Then }\quad |p(z_0+ys)|=|(1-t)p(z_0)-tp(z_0)(ys\cdot q(ys)|\leq$$ $$ \leq (1-t)|p(z_0)|+t|p(z_0)|/2=$$ $$=(1-t/2)|p(z_0)|<|p(z_0)|$$ contrary to the minimality of $|p(z_0)|.$
The long list of preliminaries was to ensure that AC hadn't been assumed somewhere.
Fix an algebraic closure $\overline{\mathbb{F}}_3$ and take $F = \varinjlim_{n = 3^k} \mathbb{F}_{3^n}\subseteq\overline{\mathbb{F}}_3$ (i.e., $F = \bigcup_{n = 3^k}\mathbb{F}_{3^n}$). Now, the polynomial $x^2 + 1$ is irreducible over $\mathbb{F}_3$ (it's quadratic, so just test to see that it has no roots). Let $i\in\mathbb{F}_9 = \mathbb{F}[x]/(x^2+1)$ be an element such that $i^2 + 1 = 0$. I claim that $i\not\in F$. To see this, we note that any element $f\in F$ lies in some finite extension $\mathbb{F}_{3^{3^k}}$, and say that no smaller extension of $\mathbb{F}_3$ contains $f$. Then $f$ satisfies a minimal polynomial with $\mathbb{F}_3$ coefficients of degree $3^k$, and as such, $f\neq i$ (because $i$'s minimal polynomial over $\mathbb{F}_3$ has degree $2$).
Best Answer
HINT: If $F$ is a field then there is a surjection from $F[x]\times\Bbb N$ onto the algebraic closure of $F$ (essentially count the roots of each polynomial, and map all the excessive points to $0$).
So it suffices to see that if $F$ is countable then $F[x]$ is countable.
[It should be noted that the axiom of choice is used here, and it is consistent with its failure that a countable field has an uncountable algebraic closure.]