Question: If $\dfrac{z-\alpha}{z+\alpha},(\alpha \in R)$ is a purely imaginary number and $|z|=2$, then find value of $\alpha$.

Now I took $\dfrac{z-\alpha}{z+\alpha}=t$ and as t is purely imaginary, and use the fact that $t+ \bar{t}=0$ and obtained the answer $\alpha = \pm2$.

But I was wondering that if there is any way to think about the answer more directly using geometry of complex numbers given that $z$ lies on a circle centered at origin having radius $2$.

Answer: 
Consider the open cover $\{X\}$. This is a finite open cover of every subset of $X$. So your modified notion would apply to every set.

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That said, there is a version of this notion which is nontrivial: what if we bound the sizes of the open sets involved? (Note that this is a fundamentally *metric*, as opposed to *topological*, idea.) Specifically, consider the following property:

> 
> For each $\epsilon>0$ there is a finite cover of $X$ consisting only of open sets with diameter $<\epsilon$.
> 
> 
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Here the diameter of an open set $U$ is the supremum of the distances between elements of that set:$$diam(U)=\sup\{d(a,b): a,b\in U\}.$$

This is no longer trivial - e.g. $\mathbb{R}$ with the usual metric does *not* have this property. But this is still very different from compactness.
If you do that, then **every** subset of $E$ of $X$ has that property. Just take the open cover $\{X\}$ of $E$. It has a finite subcover (itself).