Does there exist $\alpha \in \mathbb{R}$ and a field $F \subset \mathbb{R} $ such that $F(\alpha)=\mathbb{R}$?

No such subfield $F$ exists.

First, $\mathbb{R}$ cannot be of finite degree over $F$, else $\mathbb{C}$ would be of finite degree over $F$, and this is only possible if $F=\mathbb{R}$ or $\mathbb{C}$ by the Artin-Schreier theorem.

Thus, if such a field existed, then $\mathbb{R}$ would be of infinite degree over $F$, and $a$ would be transcendental over $F$. But then the field $F(a)$ would have many non-trivial automorphisms (for instance, the ones fixing all elements in $F$ and sending $a$ to any degree $1$ polynomial in $a$); since we know that the only field automorphism of $\mathbb{R}$ is the identity, then this is not possible.

A simpler argument pointed out by arctic tern is that neither of $a$ and $-a$ have a square root in $\mathbb{R}$, which is impossible for elements of $\mathbb{R}$.

In fact, $ \mathbb R $ does not have any subfield of finite index - this is a direct consequence of the Artin-Schreier theorem, which (in addition to other things) states that if $ C $ is an algebraically closed field and $ F $ is a proper subfield such that the degree $ [C : F] $ is finite, then $ C = F(\sqrt{-1}) $. Assuming that $ \mathbb R $ had a proper subfield $ E $ such that $ \mathbb R = E(\alpha) $ for an $ \alpha $ algebraic over $ E $ implies that the degree $ [\mathbb R : E] $, and thus $ [\mathbb C : E] $, is finite. Since $ \mathbb C $ is algebraically closed, Artin-Schreier gives the result that $ \mathbb C = E(\sqrt{-1}) $, so that $ E \subset \mathbb R \subset \mathbb C $ with $ [\mathbb C : E] = 2 $. This implies $ E = \mathbb R $, which is a contradiction.

See this writeup by Keith Conrad for more details, and a proof of the Artin-Schreier theorem.

The case when $ \alpha $ is not algebraic over $ E $ has been covered by Pierre-Guy Plamondon in his answer - I will not repeat his argument here.