Subfield of $\mathbb R$ with algebraic closure as $\mathbb C$. Does there exist a proper subfield of $\mathbb{R}$ whose algebraic closure is $\mathbb{C}$ ? 
A weaker question: Does there exist a proper subfield $F$ of $\mathbb{R}$ such that $\mathbb{R}$ is algebraic over $F$? 
 A: The two conditions you cited are equivalent, since ${\bf C}$ is the algebraic closure of ${\bf R}$
Yes, there is such a field. Just take a transcendental basis $B$ of $\bf R$ over ${\bf Q}$, and then ${\bf Q}(b)_{b\in B}\subseteq{\bf R}$ is algebraic and $\bf C$ is the algebraic closure of ${\bf Q}(b)_{b\in B}$.
I'm not sure if this is true without axiom of choice, so it might be hard to give a “concrete” example.
A: Yes, there is.
Pick you favorite algebraic number $a \notin \mathbb Q$. For example $a= \sqrt{2}$.
Consider now
$$\mathcal{F} := \{ K | K \mbox{ is a subfield of } \mathbb R , \mbox{ and } a \notin K \}$$
Now, $(\mathcal{F}, \subset)$ is a partially ordered set, which is non-empty (as $\mathbb Q \in \mathcal F$). Moreover, if $(K_{j})_j$ is a chain in $\mathcal F$, it is straightforward that $\bigcup K_j \in \mathcal{F}$.
Therefore, by Zorn Lemma,  $\mathcal{F}$ has a maximal element $\mathbb{K}$.
The maximality of $\mathbb K$ implies that $\mathbb R$ is algebraic over $\mathbb K$. Indeed, if $t \in \mathbb R$ is transcendental over $\mathbb K$ then we cannot have $a \in \mathbb K(t)$ which contradicts the maximality of $\mathbb K$. 
Indeed, if $a \in \mathbb K(t)$ then writing $a=\frac{P(t)}{Q(t)}$ we get that $P(t)-aQ(t)=0$ which means that $t$ is the root of $P-at \in \mathbb K(a)$. Then $t$ is algebraic over $\mathbb K(a)$ and $\mathbb K(a)$ is finite over $K$ which implies that $t$ is algebraic over $\mathbb K$. 
Finally, as $\mathbb R$ is algebraic over $\mathbb K$, $\mathbb C$ is also algebraic over $\mathbb K$. 
