How many fields inside $\mathbb R$? i.e. what is cardinality of $\{A \mid \ A\subset \mathbb R, A \text{ is a field} \}$?
 A: $\mathbb{R}$ has $2^{\mathfrak{c}}=2^{2^{\aleph_0}}$ subfields.
Let $\mathcal{B}$ be a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$. Then $|\mathcal{B}|=|\mathbb{R}|=\mathfrak{c}$. Every subset of $\mathcal{B}$ generates a different subfield of $\mathbb{R}$, so this gives us at least $|\mathcal{P}(\mathcal{B})|=2^{\mathfrak{c}}$ subfields. Since there are only $2^{\mathfrak{c}}$ subsets of $\mathbb{R}$ in total, there are clearly at most $2^{\mathfrak{c}}$ subfields. Thus by Schroeder–Bernstein there are exactly $2^{\mathfrak{c}}$.
A: The number $x$ of subfields of $\mathbb R$ is $x=2^{\mathfrak c}$ where ${\mathfrak c}=2^{\aleph_0}$. Here is a proof.
Choose a transcendence basis $(x_i)_{i \in I}$ of $\mathbb R$ over $\mathbb Q$, where $I$  necessarily  has cardinality $\mathfrak c$ .
a) For each subset $J\subset I$ you get a subfield $\mathbb Q ((x_j)_{j \in J})\subset \mathbb R$ and so you already get as many subfields as there are subsets $J\subset I$ , namely $2^{\mathfrak c}$.
Hence $x \geq 2^{\mathfrak c}$
b) Of course you can't have more subfields than subsets of $\mathbb R$.
Hence $x \leq 2^{\mathfrak c}$
c) From  a) , b) and Cantor-Schroeder-Bernstein, deduce $x = 2^{\mathfrak c}$
