Show that $A=\{0,1,\frac{1}{2}, \frac{2}{3}, \frac{3}{4},...,\}$ is compact.
Suppose $K=\bigcup_{\alpha\in J} U_\alpha$ is a cover for $A$.
What I want to be able to say is that since $1$ is the limit of that sequence of numbers, then since $K$ is a cover for $A$, some $U_\alpha$ must contain $1$.
And since the limit of the sequence given is $1$, then all but finitely many members of the sequence will be contained in that $U_\alpha$, then only finitely many open sets of $K$ are needed to cover the remaining members of $A$.
Your proposed solution looks good. To address your concern, $U_\alpha$ need not be an open ball centered at 1 and it still works. This is because as long as $U_\alpha$ contains 1, then $U_\alpha$ must contain some open ball around 1 by the definition of an open set.
Note: If we are allowed to use the Heine-Borel theorem, then the conclusion would immediately follow.