Proof: $A\subseteq (B\cup C)$ and $B\subseteq (A\cup C)$ then $(A - B) \subseteq C$

Proof: $A\subseteq (B\cup C)$ and $B\subseteq (A\cup C)$ then $(A - B)
\subseteq C$

How do I prove this:
Let $A, B$ and $C$ be sets, $A \subseteq (B \cup C)$ and $B \subseteq (A
\cup C)$ then $(A - B) \subseteq C$
How about this:
Let $x \in A$ and $y \in B.$
Since $A \subseteq (B \cup C)$ then $x \in (B \cup C).$
Since $B \subseteq (A \cup C)$ then $y \in (A \cup C)$
$x \in B \cup C$, so if $x \in B$, then $x \notin A - B.$
if $x \notin B$, then $x \in C.$
$y \in A \cup C$, so if $y \in A$, then $y \notin A - B.$ if $y \notin B$,
then $y \in C.$
So in both cases, $A - B \subseteq C$
Is it correct? if yes, the converse is false right?