Does an uncountable algebraically closed field of characteristic zero contain an uncountable subfield which can be embedded into $\mathbb{C}$? This is probably very simple. 
Let $k$ be an uncountable algebraically closed field of characteristic zero. Does there exist an uncountable algebraically closed subfield $k_0\subset k$ and an embedding $k_0\to \mathbb{C}$?
My feeling is that, as $k$ is uncountable, I can take a subfield $k_0\subset k$ which has the same cardinality as $\mathbb{C}$. Then, it is probably even true that $k_0\cong \mathbb{C}$. Does this work?
 A: Your reasoning doesn't work per se : $k$ might have an uncountable cardinality that is $<|\mathbb{C}|$ (for instance if the continuum hypothesis fails). 
But in this situation you can fix the issue. Here's a general situation where you don't have to make any case distinction.
$k$ is an uncountable algebraically closed field of characteristic $0$, thus by the Löwenheim-Skolem theorem, it contains a subfield $k_0$ which is algebraically closed and has cardinality $\aleph_1$ (still of characteristic $0$) 
By the Löwenheim-Skolem theorem again, $k_0$ embeds in an algebraically closed field $k_1$ of cardinality $2^{\aleph_0} = |\mathbb{C}|$, still of characteristic $0$. But then $k_1\simeq \mathbb{C}$ and we are done.
A: It is true that an algebraically closed field of characteristic 0 with the same cardinality as $\mathbb{C}$ is isomorphic to $\mathbb{C}$, so your reasoning works. This follows from the following more general fact, which can be found in many algebra books. (e.g. Proposition 9.16 in these notes I found online)

Two algebraically closed fields are isomorphic if and only if they contain a common subfield over which they have the same transcendence degree.

