Does there exist a field $K$ such that $\mathbb R \subsetneq K \subsetneq \mathbb C$?

No: the dimension of $\mathbb{C}$ as a vector space over $\mathbb{R}$ is $2$.

If $K$ is an extension field of $F$, denote by $[K:F]$ the dimension of $K$ as a vector space over $F$. Then, if $F_1\subseteq F_2\subseteq F_3$ are fields, we have $$ [F_3:F_1]=[F_3:F_2][F_2:F_1] $$ (assuming $[F_3:F_1]$ is finite then also $[F_3:F_2]$ and $[F_2:F_1]$ is finite and conversely).

---

Of course there's an elementary proof. Suppose $a+bi\in K$, $a+bi\notin\mathbb{R}$. Then $b\ne0$, so $$ i=((a+bi)-a)b^{-1}\in K $$ and therefore $K=\mathbb{C}$.

Any intermediate field between $\Bbb R$ and $\Bbb C$ is in particular an $\Bbb R$-vector subspace of $\Bbb C$. Since $\dim_{\Bbb R}\Bbb C=2$, it is either equal to $\Bbb R$ or to $\Bbb C$.

and just very simply if $z \in \mathbb{K} \subset \mathbb{C}$ but $z \notin \mathbb{R}$ then it must be of the form $a + ib$ with $a,b \in \mathbb{R}$ and $b \ne 0$. hence $i \in \mathbb{K}$ so $\mathbb{C} \subset \mathbb{R}(i) \subset \mathbb{K}$