Need help to prove $(A\cup B) \setminus (C \setminus A) = A \cup (B \setminus C)$

If you need a purely algebraic-looking proof, I would write $$ \begin{align} (A\cup B)\setminus(C\setminus A) &= (A\cup B)\setminus(C\cap A^\complement) \\&= (A\cup B)\cap (C\cap A^\complement)^\complement \\&= (A\cup B)\cap (C^\complement \cup A) \\&= A\cup(B\cap C^\complement) \\&= A\cup(B\setminus C) \end{align}$$

Assume that $x \in A \cup (B - C)$. Then ($x \in A$) OR $(x \in B$ and $x \notin C$).

If $x \in A$, then $x \in A \cup B$ and $x \notin C - A$, so $x \in (A \cup B) - (C-A)$.

If $x \in B$ and $x \notin C$, then $x \in A \cup B$ and $x \notin C - A$, so $x \in (A \cup B) - (C-A)$.

This means that $A \cup (B-C) \subset (A \cup B)-(C-A)$.

Proof of $\subseteq$:

Let $x \in (A \cup B) - (C - A)$. Then $x \in A$ or $x \in B$, and $x \notin (C-A)$. The last part means that either $x \in C \cap A$ or $x \notin C$.

Suppose $x \in C \cap A$. Then $x \in A$, so we're done as $A \subseteq A \cup (B-C)$.

Suppose $x \notin C$. But we also know that either $x \in A$ (in which case we're done) or $x \in B$ (in which case we're done, as $x \in (B-C)$).

I'll leave $\supseteq$ for you to try along similar lines.