Compactness of a Subspace of $mathbb{R}$

I am trying to determine if the following set is compact:

$$A=bigg{frac{1}{n}: nin mathbb{N}, n>0bigg}subsetmathbb{R}$$

When I consider the subspace topology induced by the standard topology of $mathbb{R}$

then I figure that the results are the discrete topology on $A$ since for any $Ysubset A$ can find an interval $(a,b)$ such that $Acap(a,b)=Y$

If I consider an open covering of elements of $mathscr{T}_A$ of $A$ consisting of every element in $A$ then no finite sub covering exists so cannot be compact.

What if I consider the set
$$A^{‘}={0}cupbigg{frac{1}{n}: nin mathbb{N}, n>0bigg}subsetmathbb{R}$$ is this set considered compact. Im guessing it is compact since it contains only sequences that converge to $0$ which is now included in the set. Consider an arbitrary open covering ${U_alpha: U_alphain mathscr{T}_{A^{‘}}, alphain I}$of $A$ If I consider a sequence ${X}_n^{infty}$ with each $X_n=frac{1}{n}in A$ where $X_nrightarrow 0$ then consider an open nbhd of $U_0$ of 0 then for $nge N$ , $X_nin U_0$ then consider the open neighborhoods $U_1,U_2, …,U_{N-1}$ of $X_1,X_2, …,X_{N-1}$ then $U_0,U_1,U_2, …,U_{N-1}$ is a finite subcovering of $A$. Is my understanding correct? Outside using the idea of the set containing all its limit points intuitively why is $A$ not compact but $A^{‘}$ is?