Claim : any rational manifold is a disjoint union of rational ball.
let's prove it for a countable rational manifold:
assume the point of $X$ are numbered $x_1,\dots,x_n...$.
Pick a neighbourhood of $x_1$ that is diffeomorphic to an open ball in $\mathbb{Q}^n$ and then pick a ball arround $x_i$ in that neighbourhood that is both open and closed (and small enough so that it is also closed in $X$) for example, pick a ball whose radius is not a square root of a rational number. Call it $F_1$.
Because $F$ is open and closed, you can easily check that $X$ is diffeomorphic to $(X-F) \cup F$. Do the same in $X - F$ for the next $x_i$ which is not in $F_1$, you get a clopen ball $F_2$ and keep going...
At then end you obtain a partition of $X$ in clopen ball $F_1, \dots F_n \dots...$ and the canonical map between $X$ and the disjoint union of the $F_i$ is a diffeomorphism. (because it is on all of the $F_i$ which form an open cover of $X$).
To break this argument and allow to recover some behavior from real differential geometry what you need is a uniform structure (or a Riemanian/metric structure for example) on your differential manifold compatible with the manifold structure which will force the existence of a "real completion". Or (almost the same) a notion of admissible cover similar to what we have in analytic rigid geometry.
EDIT : One can improve the argument to show that any to countable manifold of the same positive dimension are diffeomorphic: it suffice to choose all the clopen ball we construct to have comensurable radius in $\mathbb{Q}^n$ (for example raidus $q \pi$) this way any two balls we obtain will be diffeomorphic. If one of the manifold is formed of only a finite number of balls one can always split one of the ball in a infinite number of smaller ball using the same process with ball of radius $R/4^k$ inside the initial ball: a finite family of such balls cannot cover the initial ball by a volume argument.