Is my approach to showing $(\exists y\in B\cap C)\Rightarrow(x\in A)$ correct? Let $X$ and $Y$ be arbitrary sets. Further, let $A$ be a subset of $X,$ and let $B,C$ be subsets of $Y.$ I am currently trying to prove a statement of the form
\begin{gather}
(\exists y\in B\cap C)\Rightarrow(x\in A).
\end{gather}
My approach to the problem is to prove the (equivalent) contrapositive statement, namely,
\begin{gather}
\neg(x\in A)\Rightarrow\neg(\exists y\in B\cap C),
\end{gather}
or, equivalently,
\begin{gather}
(x\notin A)\Rightarrow(\nexists y\in B\cap C).
\end{gather}
In order to show this last statement, I show two things:
\begin{gather}
\forall y\in Y,(((x\notin A)\wedge(y\in B))\Rightarrow(y\notin C))\\
\forall y\in Y,(((x\notin A)\wedge(y\in C))\Rightarrow(y\notin B)).
\end{gather}
Unfortunately, I’m not sure whether my approach to the problem is legit.
Could you please help me see whether my approach is legal?
 A: 
or, equivalently, \begin{gather} (x\notin A)\Rightarrow(\nexists y\in
B\cap C). \tag1\end{gather} In order to show this last statement, I show
two things: \begin{gather} \forall y\in Y,(((x\notin A)\wedge(y\in
B))\Rightarrow(y\notin C))\tag2\\ \forall y\in Y,(((x\notin A)\wedge(y\in
C))\Rightarrow(y\notin B)). \tag3\end{gather}

Indeed, $$(1)\quad\equiv\quad(2)\land(3).$$ But $(1)$ needs to be rewritten as $$x∉A⇒\not\exists y\;y∈B∩C,$$ or, equivalently, $$x∉A⇒∀y\;y∉B∩C;$$ also, $(1),(2),(3)$ are not statements as they all contain a free variable $x.$

Let $X$ and $Y$ be arbitrary sets. Further, let $A$ be a subset of
$X,$ and let $B,C$ be subsets of $Y.$ I am currently trying to prove a
statement of the form \begin{gather} (\exists y\in B\cap C)\Rightarrow(x\in A). \end{gather}

Are you perhaps meaning to prove this instead? $$(\exists y{\in}Y\:y\in B\cap C)\Rightarrow(\exists x{\in}X \:x\in A)$$ Then we have
\begin{align}&\exists y{\in}Y\:y\in B\cap C\Rightarrow\exists x{\in}X \:x\in A\\
\equiv&\exists x{\in}X\;\forall y{\in}Y\;\big(y\in B\cap C\Rightarrow x\in A\big)\\
\equiv&\exists x{\in}X\;\forall y{\in}Y\;\big(y\not\in B\cap C\lor x\in A\big)\\
\equiv&\exists x{\in}X\;\forall y{\in}Y\;\big(y\not\in B\lor y\not\in C \lor x\in A\big).\tag#\end{align}
