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Definition. A subset of $\mathbb{R}$ is open if it is union of open intervals.

Proposition. Open sets are not closed under arbitrary intersections.

Let $I_{\varepsilon }=\left( -\varepsilon ,\varepsilon \right)$ for $\varepsilon >0$. Then each $I_{\varepsilon }$ is open. However, $\cap _{\varepsilon >0}I_{\varepsilon }=\left\{ 0\right\}$. Indeed, $0\in I_{\varepsilon }$ for any $\varepsilon >0$ as $-\varepsilon < 0 < \varepsilon$.

Also if $\alpha \neq 0$ then $\alpha \not\in I\left\{ \varepsilon \right\}$. It remains to show that $\left\{ 0\right\}$ is not open.

My question is: What is the $\alpha$ mean? I didn't understand last paragraph. Can you explain?

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    They are trying to prove that $I_{\epsilon} \subset \{0\}$, which is equivalent to proving that if $\alpha \in I_{\epsilon}$, then $\alpha = 0$, which is also equivalent to proving that if $\alpha \neq 0$, then $\alpha \notin I_{\epsilon}$. – 2017-01-13

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$\alpha$ is a nonzero number.

What that paragraph is trying to say, or ought to be trying to say, is that the set $\cap_{\epsilon>0} I_\epsilon$ does not contain any nonzero number, which is part of the proof that $\cap_{\epsilon>0} I_\epsilon = \{0\}$.

But, something in that last paragraph is wrong, perhaps you or someone misread or mistyped it. What it should say is "if $\alpha \ne 0$ then there exists $\epsilon > 0$ such that $\alpha \not\in I_\epsilon$". That is true using $\epsilon = |\alpha|/2$.

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    As an aside, $\epsilon$ is usually restricted to $1/n$ for some positive integer, so that one may see that even *countable* intersections of open sets are not guaranteed to be open sets. – 2017-01-13
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    @FabioSomenzi: I wouldn't agree that $\epsilon$ is "usually" restricted to $1/n$. For example, the topology on $\mathbb{R}$ can be defined using the basis of open intervals of any positive radius, or the basis of open intervals of radii $1/n$ as you say, or many other choices of basis. – 2017-01-13
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    @LeeMosher Thanks for answer. – 2017-01-13
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    @LeeMosher I should have said "when showing that the intersection of infinitely many open sets..." and maybe "often" is better than "usually." – 2017-01-13