This example is given in my book without proof so I was trying to prove it for a bit now.
I tried this below.
By Archimedian, $\exists n \in \mathbb N, \ n > x + 2$ for $x \in [-3, 3]. $Then $n + 2 > x$. This won't work as we want to show $z + 2 > x, \ z \in \mathbb Z.$
Next, the length of $[-3, 3]$ is $|-3 - 3| = |-6| = 6$ and the length of $(z - 2, z+ 2)$ is $|z - 2 - z - 2| = |-4| = 4$ and so $[-3, 3]$ doesn't fit inside $(z - 2, z+ 2)$ meaning we can't show $x \in [-3, 3] \implies \ x \in (z - 2, z+ 2) \implies \ x \in \bigcup (z - 2, z+ 2)$. But then it's not hard to prove $\mathbb R \subset (-n, n), \ n \in \mathbb N.$ By Archimedian, $\forall r \in \mathbb R \ \exists n \in \mathbb N, \ n > r$. Then $-n < - r.$ Thus $-n < -r < r < n$ and so $r \in (-n, n).$ This works even though, $|\mathbb R| > |(-n, n)| (maybe).$
Only thing I can think of is to actually construct overlapping sets $S_i$ for certain values of $z \in \mathbb Z$ s.t. $[-3, 3] \subset \bigcup S_i.$
Is there a better, elegant proof for this statement?
I don't think you realize just how minimal and simple a statement that "such and such is an open cover" is. It just means that so union of the sets contain the set.
$[-3, 3] \subset (-4,4) = (-4,-2)\cup (-3,3)\cup.... (3,4) = \cup_{z\in\{-3,-2,...,1,2\}}\subset \{(z-2,z+2)|z\in \mathbb Z\}$.
"Only thing I can think of is to actually construct overlapping sets Si for certain values"
Why wouldn't you want to do that? Doing such is trivial and obvious you do it by specifying the integers from $-3$ to $3$. Or simply claiming such a cover scans the entire reals (and so all subsets of the reals) because each interval overlaps both its successor and its predecessor. That is a complete and thorough proof.
Nothing more needs to be done. This wasn't meant to be a noodle-scratcher. It was meant to be any more than a statement like $(2,3) \subset (-5, 27)$. It's meant to be self-evident. You can prove it if you want but you really aren't expected to need to.