Proof that the theory of rationals is convex
The key idea here is that for any conjunction of equations $F\equiv u_1=v_1\wedge\ldots\wedge u_k=v_k$, the set $S_F$ is convex in the geometric sense, i.e. for any two points $p,q\in S_F$, all points between $p$ and $q$ are also in $S_F$.

Now suppose that $F\Rightarrow G$, where $G$ is a disjunction of equations $x_1=y_1\vee\ldots\vee x_n=y_n$ (note that the variables $u_i,v_i,x_i,y_i$ are not assumed to be pairwise distinct). Without loss of generality, we may assume that $G$ is minimal in the sense that for each $i$, there is a point $p_i$ in $S_F$ which satisfies $x_i=y_i$, but not $x_j=y_j$ for any $j\neq i$. Any equation for which this is not the case can simply be removed from $G$ to obtain a smaller disjunction $G'$ which still satisfies $F\Rightarrow G'$. We will show that for such a minimal $G$, $n\le 1$. 

Suppose not, i.e. $n&gt;1$. Then the line segment $l$ between (say) $p_1$ and $p_2$ is contained in $S_F$ due to (geometric) convexity. Since $p_1$ and $p_2$ cannot be the same due to their definition, $l$ has infinitely many points. Therefore there must be some $i$ such that $l\cap S_{x_i=y_i}$ has infinitely many points, and so $l\subseteq S_{x_i=y_i}$; in particular, $p_1\in S_{x_i=y_i}$, implying $i=1$, and $p_2\in S_{x_i=y_i}$, implying $i=2$ for a contradiction.