Prove that if $A \setminus B \subseteq C$ and $x \in A \setminus C$ then $x \in B$ 
Suppose A, B, and C are sets, $A \setminus B \subseteq C$, and $x$ is
  anything at all. Prove that if $x \in A \setminus C$ then $x \in B$.

Suppose $x \in A \setminus C$. 
Suppose $x \notin B$. It follows that $ x \in A\setminus C\setminus B$,  which means $x \in A \setminus B$. Since $\setminus ⊆$ it implies that $ x \in C$, which contradicts our assumption that $x \in A\setminus C $. Hence if $x \in A \setminus C $ then $x \in B$.
Is it accurate?

One more question is, wouldn't it be better if we rephrased initial statement as:

Suppose $A \setminus B \subseteq C$. Prove that if $x \in A \setminus C$ then $x \in B$.

It feels that "A,B,C are sets" and "x is anything at all" are redundant here.
 A: Since $x$ is not in $C$ and $A-B$ is a subset of $C$  we concluded that $x$ is not in $A-B$
Since $x$ is in $A$ and it is not in $A-B$ we conclude that $x$ is in $B$
A: 
It feels that "A,B,C are sets" and "x is anything at all" are redundant here.

"x is anything at all" can be rephrased as $x$ is arbitrary". Still, it is important to introduce the variables you use. One way would be to make the sets subsets of an universal set $\Omega$. The you could reword: Let $A,B,C \subset \Omega$ be sets and $x \in \Omega$ arbitrary. Prove that ...

Is it accurate?

I don't quite know what you mean by this.
Since you added the tag proof writing one suggestion would be to write
$$
\color{red}{(}A \setminus C\color{red}{)} \setminus B = A \cap C^{\complement} \cap B^{\complement}
= A \cap B^{\complement} \cap C^{\complement}
= (A \setminus B) \setminus C
$$
This makes it clear to see that
$$
[(A \setminus C) \setminus B]
\subset [A \setminus B]
$$
Otherwise your proof is perfectly fine, though I don't think you need the last sentence as it just repeats the task. An alternative would be something like "and this concludes the proof".
