Algebraic closure of $\mathbb{C}(x)$ is isomorphic to $\mathbb{C}$

If $B$ is a transcendence basis of $\mathbb C$ over $\mathbb Q$, then $B$ is uncountable, and $\mathbb C$ is an algebraic closure of $\mathbb Q(B)$.

Now $B\cup\{x\}$ is a transcendence basis of $K$ over $\mathbb Q$. Since $B$ and $B\cup\{x\}$ have the same cardinal, $K$ and $\mathbb C$ are isomorphic: they are both algebraic closures of purely transcendental extensions of $\mathbb Q$ of the same transcendence degree.

This is an immediate consequence of a famous theorem of Steinitz that uncountable algebraically closed fields are isomorphic iff they have equal characteristic and equal cardinality (a prototypical example of Morley's categoricity theorem).

This fails in the countable case, e.g. if $\mathbb A = $ algebraic numbers then, though countable, $\overline{\mathbb A(x)}\not\cong \mathbb A$ since they have unequal transcendence degree over $\mathbb Q,\:$ viz. $1$ vs. $0$, resp.