Let $X \subseteq \Bbb Q^2$. Suppose each continuous function $f:X \to \Bbb R^2$ is bounded. Then $X$ is finite. 
True or false: Let $X \subseteq \Bbb Q^2$. Suppose each continuous function $f:X \to \Bbb R^2$ is bounded. Then $X$ is finite.

Now it will be compact for sure just by using distance function.
Now what can we do? 

In a metric space $(X,d)$, a subspace $Y \subseteq X$ has the property that all continuous $f: Y \to \Bbb R$ are bounded ($Y$ is then called pseudocompact) iff $Y$ is a compact subspace of $X$. See here e.g. And $X=\mathbb{Q}^2$ has many infinite compact subsets, like any convergent sequence together with its limit.
