Characterization of a subfield $K \varsubsetneq \mathbb {C}$ and $x\in \mathbb{R}$ 
Characterize $x \in \mathbb R$ such that there exist a subfield $K \varsubsetneq \mathbb C$ such that $K(x) = \mathbb C$.

-All subfields $K$ of $\mathbb{C}$ contain $\mathbb Q$, then all $x\notin \mathbb Q$
-If there exist $K$, it is uncountable. In fact if $K(x)= \mathbb {C}$ is countable, which is absurd.
But this is not enough to characterize all the x..,
Thank you in advance,
 A: There is no such number！ Suppose otherwise there is such a number $x\in \mathbb{R}$ with the property that there exists a proper subfield $K$ of $\mathbb{C}$ such that $K(x)=\mathbb{C}$. If $x$ is algebraic over $K$, then $\mathbb{C}$ is a finite extension of $K$. By Artin-Schreier theorem we know $K$ is just $\mathbb{R}$ which is impossible since $x\in \mathbb{R}$. If $x$ is transcendental over $K$, then $K(x)$ is not algebraically closed which is absurd since $K(x)=\mathbb{C}$.
There is a point to be make precise. In fact, by Artin Schreier theorem, if $x$ is algebraic over $K$, we actually get that $K$ is real closed. Therefore $K$ can be defined as the field of real numbers. So in this sense there is no such number $x$. 
But if $\mathbb{R}$ is already choosed in $\mathbb{C}$, then the answer to the question is that such numbers are precisely real transcendental numbers. 
First, let's show any real transcendental number $x$ can do. Choose a transcendental base $B$ of $\mathbb{R}$ over $\mathbb{Q}$. We may assume $B$ contains $x$. Consider the set $B'$ which is obtained from $B$ by repacing $x$ by $x+i$ and the other elements left unchanged, where $i\in \mathbb{C}$ such that $i^2=-1$. Then $B'$ is also a transcendental base of $\mathbb{C}$ over $\mathbb{Q}$. So there exists an automorphism $\alpha : \mathbb{C} \rightarrow \mathbb{C}$ such that $\alpha(x)=x+i$ and $\alpha(y)=y$ for any $x\neq y\in B$. Take $K=\alpha(\mathbb{R})$. Thus $i\notin K$ since $i\notin \mathbb{R}$ and $[\mathbb{C}:K]=2$. Now we know $x\notin K$ for otherwise $i=(x+i)-x \in K$. Therefore we know $K(x)=\mathbb{C}$.
Now, we are left only to show that any real algebaric number $x$ can not do. Suppose otherwise there is a real algebaric number $x$ with the desired property. Since $x$ is algebraic number we know the extension $\mathbb{C}/K$ is finite. Hence by Artin Schreier theorem $K$ is real closed. Thus we can find an isomorphism $\beta : \mathbb{R}\rightarrow K$. So all real conjugates including $x$ itself lies in $K$ which contradict to the fact $x\notin K$.
A: Denote by $E$ the set of $x \in \mathbb{R}$ satisfying your conditions.


*

*If $x \in \mathbb{R}$ is transcendental over $\mathbb{Q}$ or algebraic not totally real, then $x \in E$.



Take $y=i \pi$ if $x$ is transcendental or $y \notin \mathbb{R}$ conjugate to $x$ if $x$ is algebraic not totally real. Using Zorn's lemma you can find an automorphism $\varphi$ of $\mathbb{C}$ such that $\phi(y)=x$. Take $K:=\phi(\mathbb{R})$. Then $K[x] = \varphi(\mathbb{R}[y])=\mathbb{C}$



*

*If $x$ is algebraic totally real, then $x \notin E$.



Unfortunately I don't know how to do it without Artin-Schreier. Recall that Artin-Schreier states that if $K \subset \mathbb{C}$ is such that $[\mathbb{C}:K]<\infty$ then there exists an automorphism $\varphi$ of $\mathbb{C}$ inducing an isomorphism $\varphi : \mathbb{R} \xrightarrow{\sim} K$.

A: Building on the ideas in the comment, let us prove the following statement:

Suppose that $K$ is a field and $t\notin K$ is transcendental over $K$, then $K(t)$ is not algebraically closed.

In particular, let us see that $\sqrt t\notin K(t)$, so $x^2-t$ doesn't have a root in $K(t)$. Assume otherwise, then $\sqrt t\notin K$ as well, because $t\notin K$. So by definition there is a rational function $h(x)=\frac{f(x)}{g(x)}$ such that $h^2(x)=x$.
We have that $f^2(x)=x\cdot g^2(x)$. Let $n=\deg f$ and $m=\deg g$. Then we have $n^2=m^2+1$, so we have that $(n+m)(n-m)=1$. Since both are non-negative integers this means that $n=1$ and $m=0$, and therefore $f(x)=ax+b$, and $g(x)=c$ for some $a,b,c\in K$.
Therefore $\deg h=\deg f$ and so $\deg h^2=2\neq \deg x=1$. $\quad\square$

Now suppose that $x\in\Bbb R$, and $K$ is a proper subfield of $\Bbb C$ such that $K(x)=\Bbb C$, in particular $K(x)$ is algebraically closed. This is impossible if $x\notin K$, and therefore we have a contradiction since $K(x)\subseteq K\neq\Bbb C$.
