I'm having some trouble proving this theorem. Here's what I have so far.
Theorem. For $a, b \in \mathbb{Z}$, $a \mathbb{Z} \subset b \mathbb{Z} \iff b \mid a$.
Proof. Fix two arbitrary integers, $a$ and $b$, and define the sets $a\mathbb{Z}$ and $b\mathbb{Z}$ containing all of their multiples.
Step 1: $a\mathbb{Z} \subset b \mathbb{Z} \implies b \mid a$.
First, suppose that $a \mathbb{Z} \subset b\mathbb{Z}$. Let $z$ be an arbitrary integer, so $y = a_iz_i \in a \mathbb{Z}$, which implies that $y = b_a z_a \in b \mathbb{Z}$, so w have $a_i z_i = b_a z_a$.
The reason for the subscripts is to hopefully come to some conclusion about the divisibility of $b$. Euclid's Lemma is the first fact that comes to mind, but clearly neither of these terms are prime. The argument, conceptually, is something to the effect of this: the term on the left clearly must divide either $b_a$ or $z_a$. If it divides $z_a$, we can continue to move factors toward the 'b' term so that we allow $z_i$ and $z_a$ to be relatively prime, which then requires that $a$ divide $b$ to maintain equality. I'm unsure on whether this is a satisfactory argument, and perhaps less sure on how to formalize it if it is.
Step 2: $b \mid a \implies a\mathbb{Z} \subset b\mathbb{Z}$.
Suppose $b \mid a$, so there exists some $c \in \mathbb{Z}$ such that $bc = a$. Let $az$ be an arbitrary element of $a\mathbb{Z}$. We have $az = (bc)z = b(cz)$, where $cz \in \mathbb{Z}$ by closure. Thus, $az \in b\mathbb{Z}$.
The second component of this seems far easier, assuming these are valid steps, but I still am struggling a bit with the first step. Any helpful comments would be greatly appreciated.
You're going a wrong path for step 1.
Suppose $a\mathbb{Z}\subset b\mathbb{Z}$; in particular $a\in b\mathbb{Z}$, so $a=bk$ for some integer $k$, which implies $b\mid a$.
Your step 2 is correct.