Proving a Subset - $(A∪B)∩C⊆A∪(B∩C).$ I'm having a hard time with my subset proof. I think I'm skipping over some steps.

Let $A$, $B$, and $C$ be sets. Prove that $(A ∪ B) ∩ C ⊆ A ∪ (B ∩ C).$

Theorem: $(A ∪ B) ∩ C ⊆ A ∪ (B ∩ C).$
Proof:
Let $x ∈ (A ∪ B) ∩ C$
Assume: 
If $x ∈ A$ or $x ∈ B$, then $x ∈ C$; since $(A ∪ B) ∩ C$
If $x ∈ A$, then $x ∈ C$
If $x ∈ B$, then $x ∈ C$
$∴ x ∈ A ∪ (B ∩ C)$

As Cornman notes, it seems you are having an issue primarily with sorting what you are given and how exactly that is meant to be used in what you are trying to prove. You write, "If $x\in A$ or $x\in B$, then $x\in C$," but you do not have an implication there. The overall implication you are trying to show is that
$$
x\in(A\cup B)\cap C\implies x\in A\cup(B\cap C);\tag{1}
$$
that is, you are trying to show that $(A\cup B)\cap C\subseteq A\cup(B\cap C)$ [recall that the definition of subset involves an implication, namely $A\subseteq B=\{x : x\in A\implies x\in B\}$]. 
To that end, you want to start by assuming $x\in(A\cup B)\cap C$, and then show that this assumption implies that $x\in A\cup(B\cap C)$. Note that if $(A\cup B)\cap C$ is somehow empty, then $(1)$ is vacuously true. So let's start by assuming $x\in(A\cup B)\cap C$ and see if we can establish that this necessarily implies that $x\in A\cup(B\cap C)$, as desired.
Here is one way of looking at it:
$$
\begin{align*}
x\in(A\cup B)\cap C
&\implies x\in A\cup B\ \text{and}\ x\in C & \text{(def. of $\cap$)}\\[0.5em]
&\implies (x\in A\ \text{or}\ x\in B)\ \text{and}\ x\in C & \text{(def. of $\cup$)}\\[0.5em]
&\implies (x\in A\ \text{and}\ x\in C)\ \text{or}\ (x\in B\ \text{and}\ x\in C) & \text{(case analysis/distrib.)}\\[0.5em]
&\implies (x\in A)\ \text{or}\ (x\in B\ \text{and}\ x\in C) & (\text{since}\ A\cap C\subseteq A)\\[0.5em]
&\implies x\in A\ \text{or}\ x\in B\cap C & \text{(def. of $\cap$)}\\[0.5em]
&\implies x\in A\cup(B\cap C) & \text{(def. of $\cup$)}
\end{align*}
$$
Ultimately, you should definitely know distributivity rules concerning $\cap$ and $\cup$, which tells us that
$$
(A\cup B)\cap C=(A\cap C)\cup(B\cap C),
$$
whereby it should be clear that we have
$$
(A\cup B)\cap C\subseteq A\cup(B\cap C)
$$
as a natural result since $A\cup(B\cap C)$ is certainly "bigger" than $(A\cap C)\cup(B\cap C)$. 
