Why is the theory of reals convex? I am learning about first-order theories, and have read that the convexity of a theory is defined as follows:

A $\Sigma$-theory $T$ is convex if for every quantifier-free conjunctive $\Sigma$-formula $F$ and for every disjunction $\bigvee_{i=1}^n (s_i=t_i)$, if
  $$F\implies\bigvee_{i=1}^n(s_i=t_i)$$
  then for some $i\in \{1,2,\dots,n\}$,
  $$F\implies s_i=t_i$$

I have also read that the theory of integers, $T_{\mathbb Z}$, is not convex. A common variant of a counterexample I have seen is: let $F: 0\le x\wedge x\le 1$. Then, $F\implies x=0\vee x=1$, but $F\nRightarrow x=0$ and $F\nRightarrow x=1$.
But why is the theory of reals, $T_{\mathbb R}$ (with $\Sigma_{\mathbb R}=\{0,1,+,-,\cdot,=,\ge\}$) , convex? I have seen this claim, but have not seen any proof so far. I came up with the following example:
Let $F: x^2 - 5x + 6 = 0$. Then, we have,
$$F\implies x=2\vee x=3$$
but
$$F\nRightarrow x=2$$
and
$$F\nRightarrow x=3$$
This suggests that $T_{\mathbb R}$ should be non-convex as well.
My questions are:


*

*What am I missing in the above example?

*How can I prove that $T_{\mathbb R}$ is convex?


I think you need to specify what you include in “the theory of the reals”. I see here: 
https://courses.engr.illinois.edu/cs576/sp2017/readings/31-may-02/oppen-convex.pdf 
an argument that the theory of the rationals under sum and $\leq$ is convex, while the theory of the reals under multiplication is not; your proposed counterexample needs both addition and multiplication.
