Take a compact Cantor set $K \subseteq \mathbb{R}$ of Hausdorff dimension zero. Actually we need all cartesian powers $K^n$ of dimension zero as well. The field $\mathbb{Q}(K)$ generated by it is uncountable, but still of Hausdorff dimension zero, so it is a proper subfield.

**edit**

That field consists of the values of rational functions $w(x_1,\dots,x_n)$ of many variables with rational coefficients, where the variables range over $K$. There are countably many such things, so you just have to show any one of them has dimension zero. The domain of any such $w$ (that is, the set where the denominator does not vanish) consists of an increasing countable union $\bigcup_k A_k$ of sets where the gradient is bounded, so that $w$ is Lipschitz continuous on each $A_k$. So the image of $w$ on $K^n$ is again a countable union of sets of dimension zero.