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? 
 A: Hint: Consider $X = \{(1,0), (1/2,0), (1/3,0), ..., (0,0)\}$. What can we say about the behaviour of $f(x)$ as $x\to (0,0)$?
A: 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.
A: Any subspace $X\subset \Bbb R^2$ is compact iff $X$ is closed in $\Bbb R^2$ and bounded with respect to the usual metric.
A continuous image of a compact space is compact, so if $X$ is compact and $f:X\to \Bbb R^2$ is continuous then the image $f(X)$ is compact, hence $f(X)$ is bounded.
For example $X=\{(0,0)\}\cup \{(1/n,0):n\in \Bbb N\}$ is closed and bounded.
Remarks. (i). If $X$ is an unbounded subset of $\Bbb R^2$  then id$_X:X\to \Bbb R^2$ is continuous and has unbounded range. (ii). If $X$ is a non-closed subset of $\Bbb R^2,$ take $p\in \overline X\setminus X$... And with the usual distance-function (metric) $d$, let $f(x)= (1/d(x,p),\,0)$ for $x\in X $. Then $f:X\to \Bbb R^2$ is continuous and unbounded.
