Let $A\subset(0,\infty)$. Now $X(A)\subset\mathbb{R}^2$ will be a sum of closed intervals connecting points $(0,-1)$ and $(a,\frac{1}{a})$, $a\in A$.
I am asked to prove $X(A)$ is compact $\iff A$ is compact $\iff X(A)$ is closed. The topology is natural.
I find it very hard to write anything beside saying it's obvious. How would you go from talking about line segments to talking about points?
Let me organize things a bit for you so it's clear what you have to do:
It's easy to see that $A$ is bounded $\iff$ $X(A)$ is bounded. Therefore you have prove only that $X(A)$ is closed $\iff$ $A$ is closed and $X(A)$ is closed $\implies$ $A$ is bounded.
Clearly $X(A)$ is closed $\implies$ $A$ is closed because $A$ is an intersection of $X(A)$ with a closed set (which one?).
The last implication can be proven by taking the contrapositive: $A$ is unbounded $\implies$ $X(A)$ is not closed, do you see this?
What remains is $A$ closed $\implies$ $X(A)$ closed, can you prove that?