See, I am told that $(0,1)$ is not compact as a subspace of $\mathbb{R}$. Question is, how do I conclude that?

The hint says 

> 
> $(\epsilon,1)\_{\epsilon>0}$ does not have a finite subcover.
> 
> 
> 

But I am unsure why $(\epsilon,1)$ does not have a finite (open) subcover for $(0,1)$. The right end point is fixed to $1$ so all I can do is move around $\epsilon$. So some $\epsilon$ as close as possible to zero $(\epsilon,1)$ where $\epsilon \to 0$ alone would be an open cover of $(0,1)$.

Any other $\epsilon>0$ I choose would be redundant. $(1/2,1),(1/60,1),(8/7,1)...$ whatever. These are already covered by $(\epsilon,1)$ where $\epsilon \to 1$. So that means, the open cover by $(\epsilon,1)$ is finite on its own to start with. namely, just one open subset $(\epsilon,1)$ of $\mathbb{R}$ covers $(0,1)$.

I can come up with loads of other open covers, sure, but as I said, any other open cover apart from $\epsilon \to 0$ can be removed and still be an open cover of $(0,1)$. e.g. $\{(\epsilon,1)\_{\epsilon \to 0},(23/81,1)\}$ is an open cover of $(0,1)$ but I can get rid of $(23/81)$ if I want to since that still leaves me with an open cover of $(0,1)$. So the same for any infinite open cover $\{(\epsilon,1)\_{\epsilon \to 1},(1/2,1),(2/5,1),(1/3,1),...\}$ I can get rid of everything that follows $(\epsilon,1)$ and be left with an open cover i.e. finite subcover.

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I need some clarification here and most probably; I am mistaken with something so blatantly obvious to those with experience. Think I am a super stupid human being and explain from possible the $1+1=2$s of this bizarre world of compactness if necessary. I think someone needs to fix some basic but esoteric notion for this one.