You're on the right track. If we consider $X=\left\{\frac1n:n\in\Bbb N^+\right\}$ in the discrete topology, then we can endow it with the metric $d:X\times X\to\Bbb R$ given by $$d(x,y)=\begin{cases}0 & x=y\\1 & \text{otherwise,}\end{cases}$$ which does indeed induce the discrete topology on $X$ (it's called the discrete metric for this reason). Then $X$ is certainly bounded, as any ball of radius greater than $1$ necessarily includes the whole set, and is certainly closed in itself (as all spaces are). However, it is not compact, since the open cover by singletons admits no finite subcover, as you've observed. More generally, any infinite discrete space admits a proper subspace that is closed and bounded, but not compact (delete any point).
We could come to the same conclusions if we considered $X$ as a space under the metric $$\rho(x,y)=|x-y|.$$ Indeed, $\rho$ induces the discrete topology on $X$, as well, and we similarly find that $X$ is bounded under $\rho$.
The kicker, here, is the boundedness. You need to specify a metric, or some other convention to determine boundedness, not just a topology. For example, $\Bbb Z$ considered as a subspace of $\Bbb R$ is indeed discrete, but while it is bounded in the discrete metric, it is not bounded in the standard metric on $\Bbb R$.
More simply, note that $(x,y)\mapsto xy$ is a continuous function $\Bbb R^2\to\Bbb R,$ and that $\bigl\{(x,y)\in\Bbb R^2:xy=1\bigr\}$ is the preimage of the closed set $\{1\},$ so is closed by continuity.
Best Answer
Hint: Let $(x_n,y_n)$ be convergent sequence of elements such that $(x_n,y_n) \in Graph$. Prove that $\lim (x_n,y_n) \in Graph$