Can it be proven in ZFC that there exists a field with $\aleph_1$ subfields? Can it be proven in ZFC that there exists a field with $\aleph_1$ subfields? I am not asking whether such a field is definable in ZFC, I am asking whether the statement "there exists a field $K$ with $\aleph_1$ subfields", after a suitable translation into first-order logic, can be proven in ZFC without using the continuum hypothesis.

No; indeed, any field $K$ with uncountably many subfields must have at least $2^{\aleph_0}$ subfields, so your statement is equivalent to CH.  If $K$ has infinite transcendence degree over the prime field, you can get $2^{\aleph_0}$ subfields by taking subfields generated by arbitrary subsets of an infinite algebraically independent set.  If $K$ has finite transcendence degree of the prime field, then in particular it is countable.  Now observe that the set of (characteristic functions of) subfields of $K$ is a closed subset of $\{0,1\}^K$ (in the product topology).  Since $K$ is countable, $\{0,1\}^K$ is just a Cantor set, so by the Cantor-Bendixson theorem, every uncountably closed subset of it has cardinality $2^{\aleph_0}$.
