Is there a subfield $F$ of $\Bbb R$ such that there is an embedding $F(x) \hookrightarrow F$?
I think I've found an example, thanks to this answer. We pick a subset $\{x_i\}_{i\in \mathbb{N}}$ of pairwise distinct elements of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$ (this is possible since a transcendence basis of $\Bbb R$ over $\Bbb Q$ has cardinality $2^{\aleph_0}$).
Then the subfield $F = \Bbb Q(x_1, \dots, x_n, \dots)$ satisfies $F \cong F(X) \cong F(X,Y) \cong \cdots$ (as fields), so in particular $F(X)$ embeds in $F$.