Find a closed, bounded subset $A$ of $\Bbb Q$ and a continuous function $f : A →\Bbb R$ such that $f$ is not bounded
Note: $\Bbb Q$ is the set of all rationals.
My Solution:
$A=\{x:x\in\Bbb Q, 1\leq x\leq2\}$. Clearly $A$ is bounded below by $1$ and bounded above by $2$. Also $A$ is closed since it contains its limit points. Define $f : A →\Bbb R$ such that $f(x)=\frac1{x-1}$. $f(x)$ is not bounded at $x=1$ but I am unsure about its continuity at $x=1$. Does the right continuity at $x=1$ alone ensure the function is continuous?
What is more important is the question that how does this not violate the theorem that the image of a compact set is compact under a continuous function. Since $A$ is compact (any open cover of $A$ has a finite subcover), the images of $A$ under $f$ must also form a compact set which implies that the image set is closed and bounded, contradiction.
Unfortunately, that doesn't work, since $f$ is not defined at $1$ or at $2.$ However, you have the right general idea. Instead, you can pick some irrational number $\beta$ with $1<\beta<2,$ and let $f(x)=\frac1{x-\beta}.$
Alternately, proceed more or less as you have, but instead, let $A:=\{x\in\Bbb Q:\alpha\le x\le\beta\}$ for some irrational $\alpha,\beta$ with $\alpha<\beta,$ then put $f(x)=\frac1{(x-\alpha)(x-\beta)}.$ $A$ turns out to be closed, and clearly bounded, and the rest is fairly straightforward.
As a side note, $\Bbb Q$ (with the usual metric) is a good example of a metric space in which "closed and bounded" need not imply "compact," the demonstration of which is (I suspect) more or less the point of this exercise.