Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I have shown that the collections of sets $$S_1=\left \{\left (\frac{x}{2}, \frac{3x}{2}\right ): 0<x<1\right \}, \ \ \ \ \ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}$$ are open covers of $A=\left \{\frac{1}{n} : n\in \mathbb{N}\right \}\subset \mathbb{R}$.
To check if they have a finite subcover, we have to check if $S_1$ and $S_2$ respectively, have a finite subset which is also a cover of $A$, right?
For example, we have that $\displaystyle{\tilde{S}_1=\left \{\left (\frac{x}{4}, \frac{3x}{4}\right ): 0<x<1\right \}}$ is a subset of $S_1$, isn't it?
This is not a cover of $A$, since for $n=1$ the element $a$ is not in $\tilde{S}_1$. The same holds also when we consider a subset of the form $\left \{\left (\frac{x}{i}, \frac{3x}{i}\right ): 0<x<1\right \}$ with $i>2$.
Does this mean that $S_1$ has no finite subcover?
Let $F$ be a non-empty finite subset of $S_1$. Then $F=\{(x/2,3x/2): x\in G\}$ where $G$ is a finite non-empty subset of $(0,1).$ Let $y=\min \{x/2:x\in G\}.$
Then $y>0.$ If $z$ is any member of $ \cup F$ then for some $x\in G$ we have $z\in (x/2,3x/2)$ so $z>x/2\geq y.$
Take $n\in \Bbb N$ with $1/n<y.$ Then $1/n$ is less than every member of $ \cup F.$ So $1/n\not \in \cup F$. So $F$ is not a cover of $A. $