Arbitrary $\epsilon$ in Analysis Proof I searched the Math stack, but I was unsuccessful in obtaining an adequate response. Please feel free to share a link if this is a repeat AND there is a good response.
I am encountering picking or fixing an $\epsilon$ for the following proof in the Exterior Measure section by Stein and Shakarchi (2009).
We are trying to prove 
$$|Q|\leq\sum^\infty_j|Q_j|.$$
The authors uses the fact that $Q$ is a compact set and chooses an open cube arbitrarily close to the original closed cube we started with, in conjunction with the Heine-Borel theorem and claim:
$$|Q|\leq\sum^N_j(1+\epsilon)|Q_j|\leq\sum^\infty_j(1+\epsilon)|Q_j|.$$
Then conclude, $\epsilon$ was chosen arbitrary, so it shows what we want to prove.
Why is this true? Is this because in Analysis, $a_1=a_2$ means $|a_1-a_2|<\epsilon$ for some arbitrarily small $\epsilon$?
More importantly, why do we use this trick? The pragmatic reason I see here is that we are starting with an arbitrary covering $(Q_j)$ that covers a closed cube $Q$. To get around it, we choose an aribtarily close-looking cube $(S_j)$ that each contains $Q_j^s$, so in order to do this the authors establish:
$$|S_j|\leq(1+\epsilon)|Q_j|$$
It seems to be without the epsilon such construction would not be possible. So is this the real reason why we say fix an arbitrary $\epsilon$ and then eventually make that go to zero for the inequality to ultimately hold?
Reference:
$\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.
 A: Answering the "why do we use this trick?" question:
The authors want to use an open covering so they can reduce the countable covering to a finite one (by compactness of $E$). So, they are enclosing each $Q_j$ in an open cube $S_j$.
If $Q_j$ is not open, and $S_j$ is an open cube containing $Q_j$, then $S_j$ cannot have the same volume as $Q_j$; it must be strictly larger.
For example, in one dimension, the "cube" (interval) $[a,b]$ has length $b-a$, but any open interval containing $[a,b]$ has length strictly greater than $b-a$. However, we can find an open interval with length as close to $b-a$ as desired, for example, we can make it have length $(b-a)(1+\epsilon)$ by choosing the interval $(a - \delta, b + \delta)$, where $\delta = (b-a)\epsilon/2$.
Similarly, in $n$ dimensions, you can do something quite similar with a cube of volume $V$: you can find an open cube containing the original cube, with volume $V(1+\epsilon)$, or with volume $V + \epsilon$, or any other function of $V$ and $\epsilon$ that is slightly larger than $V$ and converges to $V$ as $\epsilon \to 0$.
Btw, I'm not sure why the authors aren't using open $Q_j$'s to begin with. Most authors define outer measure in terms of coverings by countable sequences of open intervals / cubes in part because they can be reduced to a finite subcover when covering a compact set.
A: 
Is this because in Analysis, $a_1=a_2$ means $|a_1-a_2|<\epsilon$ for some arbitrarily small $\epsilon$?

What's in fact true is that $|a_1-a_2|<\epsilon$ for all $\epsilon>0$ means $|a_1-a_2|=0,$ which of course implies $a_1=a_2.$ The converse is even easier to see.

More importantly, why do we use this trick?

I don't see a trick here. I simply see it as an application of the triangle inequality.
