Prove or Disprove :
Let $X \subset \mathbb{Q}^2$. Suppose that each continuous function $f:X\to \mathbb{R}^2$ is bounded. Then $X$ is necessarily finite.
I think this statement is wrong as if we know that every continuous function takes compact sets to compact sets. Now every compact set is bounded, so its image is compact which implies it is bounded. So we have to construct a compact set which is not finite ......but I can't figure out any example .....
This is false. It is possible for every continuous function on $X$ to be bounded even when $X$ is infinite. For example take $X=\{0,1,\frac1 2,\frac 1 3,...\} \times \{0,1,\frac 1 2,\frac 1 3,...\}$. By compactness of $X$ every continuous function on $X$ is bounded.
I'll just prove the statement for $\mathbb{Q}$ and $\mathbb{R}$, as it is basically the same problem. Suppose $X$ is infinite. Choose a countable increasing subset (or, if one doesn't exist, choose a decreasing countable subset): $$\{a_1, a_2, \dots \}.$$ Define $f(a_n)=n$ and "connect the dots" to get a continuous function.
