How does one prove (A - B) - C ⊆ (A - C) - (B - C) When proving this I'm not sure how to 'take out' the C on the RHS of the equation.
The LHS is
(x ∈ A) ∧ !(x ∈ B) ∧ !(x ∈ C)
The RHS is
(x ∈ A) ∧ !(x ∈ C) ∧ !(x ∈ B) ∧ !(x ∈ C)
How does how prove LHS is a subset of RHS?

Just show that each $x\in(A\setminus B)\setminus C$ belongs to $(A\setminus C)\setminus(B\setminus C)$.
Suppose that $x\in(A\setminus B)\setminus C$; then $x\in A\setminus B$, and $x\notin C$. Since $x\in A\setminus B$, we know further that $x\in A$ and $x\notin B$. 
Now put the pieces back together. First, $x\in A$ and $x\notin C$, so $x\in A\setminus C$. Moreover, $x\notin B$, so certainly $x\notin B\setminus C$, since $B\setminus C$ is a subset of $B$. But that means that $x\in A\setminus C$ and $x\notin B\setminus C$, which is exactly what’s required to say that $x\in(A\setminus C)\setminus(B\setminus C)$.
Since $x$ was an arbitrary element of $(A\setminus B)\setminus C$, this shows that every element of $(A\setminus B)\setminus C$ belongs to $(A\setminus C)\setminus(B\setminus C)$ and hence that $(A\setminus B)\setminus C\subseteq(A\setminus C)\setminus(B\setminus C)$.
(I call this approach element-chasing. It’s one of the most straightforward ways to prove that one set is a subset of another.)
