Prove that $\vert \bigcup_{i \in I} A_i \vert \le \vert I \vert\alpha$ (with context). Let $\alpha$ be a fixed cardinal number and suppose that for every $i \in I$, $A_i$ is a set with $\vert A_i \vert = \alpha$. Then $\vert \bigcup_{i \in I} A_i \vert \le \vert I \vert\alpha$.
Edited proof. Please verify it and provide feedback: 
First, consider the case that all $A_i$ are disjoint. Then $\vert \bigcup_{i \in I} A_i \vert$ must be some integer multiple of $\alpha$; in fact, it is exactly equal to $\vert I \vert\alpha$ because for every $i \in I$, $A_i$ appends an additional $\alpha$ elements to $\vert \bigcup_{i \in I} A_i \vert$. Now, consider the case where not all $A_i$ are disjoint. Then, there exists at least one pair of sets $A_i, A_j$ where $i, j \in I$ that are not disjoint. $\vert A_i \cup A_j\vert = 2\alpha - \beta$, where $\beta = \vert A_i \cap A_j\vert.$..
How would I rigorously show that this subtraction of $\beta$ would then extend to the union $\bigcup_{i \in I}A_i$?
Note: I'm interpreting the product of cardinal numbers to be a literal multiplication of natural numbers, and not the cardinality of the cartesian product of their respective, original sets.
 A: The first part seems okay as long as you are assuming that all of the cardinal numbers are finite. However I'm not sure where $\beta$ comes from, or how it's okay for it to only depend on two elements of the family $(A_i)_{i\in I}$. Here's an argument that works for arbitrary cardinals.
Since $|\cup_{i\in I}A_i|\leq |\sqcup_{i\in I}A_i|$ (see below), where $\sqcup$ denotes a disjoint union, it suffices to prove the claim whenever $(A_i)_{i\in I}$ are pairwise disjoint. Let $A$ be a set such that $|A|=\alpha$.
From $|A_i|\leq\alpha$ we get an injection $f_i:A_i\to A$, for each $i\in I$. (Caution: To select $f_i$ for all $i\in I$, we used the axiom of choice.) Define
$f:\cup_{i\in I}A_i\to I\times A$ by
$$
f(a) = (i, f_i(a)), \quad \text{where}\ i\in I\ \text{is such that}\ a\in A_i.
$$
This map is well-defined because $(A_i)_{i\in I}$ are pairwise disjoint. We also have that $f$ is injective. Indeed, if $f(a)=f(b)$, then $a,b\in A_i$ for some $i\in I$ and $f_i(a)=f_i(b)$, from which we obtain $a=b$ by the injectivity of $f_i$. Therefore
$$
|\cup_{i\in I}A_i | \leq |I\times A| = |I||A|=|I|\alpha.
$$


If $(A_i)_{i\in I}$ is a family of sets, then 
  $|\cup_{i\in I}A_i|\leq |\sqcup_{i\in I}A_i|$.

Proof:
This is clearly true if $I=\emptyset$, and if some $A_{i_0}=\emptyset$, then $\cup_{i\in I}A_i=\cup_{i\in I\setminus\{i_0\}}A_i$ and $\sqcup_{i\in I}A_i=\sqcup_{i\in I\setminus\{i_0\}}A_i$ shows that it is enough to prove the result for $(A_i)_{i\in I\setminus\{i_0\}}$.
Thus we may assume $(A_i)_{i\in I}$ is a nonempty family of nonempty sets.
Recall $\sqcup_{i\in I}A_i=\cup_{i\in I}B_i$, where $B_i:=\{i\}\times A_i$. By the axiom of choice, there exists a function $c:\cup_{i\in I}A_i\to I$ such that $a\in A_{c(a)}$. Then define an injection $f:\cup_{i\in I}A_i\to\sqcup_{i\in I}A_i$ by
$$
f(a) = (c(a),a).
$$
