Is my approach to showing $(x\in A)\Rightarrow(\exists y\in B\cap C)$ correct?

$$(x\in A)\Rightarrow(\exists y\in B\cap C).\tag1$$

Grammatical correction (duplicating the $y):$ $$x\in A\Rightarrow\exists y\;y\in B\cap C.$$

These are all equivalent to $(1):$ \begin{gather} \exists y\;\big(x\in A\Rightarrow y\in B\cap C\big)\\\exists y\;\big(x\not\in A\lor y\in B\cap C\big)\\\exists y\;\big(x\not\in A\lor (y\in B\land y\in C)\big).\tag#\end{gather}

My approach is to prove the (equivalent) contrapositive statement \begin{gather} (\nexists y\in B\cap C)\Rightarrow(x\notin A).\end{gather}

Grammatical correction (the parentheses here are only for disambiguation though): $$\big(\nexists y\;y\in B\cap C\big)\Rightarrow x\notin A.\tag1$$

Note that this is equivalent to (here, the parentheses are required) $$\exists y \;\big(y\not\in B\cap C\Rightarrow x\notin A\big).$$

All the above formulae are equivalent to one another; taking contrapositive doesn't appear to offer any advantage.

In order to show this last statement, I show two things: \begin{gather} \forall y\in Y,((y\notin C)\Rightarrow(x\notin A))\tag2\\ \forall y\in Y,((y\notin B)\Rightarrow(x\notin A))\tag3 \end{gather}

\begin{gather} \exists y{\in}Y\;\big(y\notin C\Rightarrow x\notin A\big)\tag4\\ \exists y{\in}Y\;\big(y\notin B\Rightarrow x\notin A\big)\tag5 \end{gather}

If we ignore the quantifiers, then $\left[(2)\land(3)\right]$ is indeed equivalent to $(1).$

However, with quantifiers, things get more complicated:

• $\left[(2)\land(3)\right]$ implies $(1)$
• $(1)$ does not imply $\left[(2)\land(3)\right]$
• $(1)$ implies $\left[(4)\land(5)\right]$
• $\left[(4)\land(5)\right]$ does not imply $(1)$

In summary, it is probably best to stick to using $(\#).$

Let $\overline{B}$ be the complement of $B$ in $Y$, and a similar definition for $\overline{C}$. Then the statements \begin{gather} \forall y\in Y,((y\notin C)\Rightarrow(x\notin A))\tag2\\ \forall y\in Y,((y\notin B)\Rightarrow(x\notin A))\tag3 \end{gather} are equivalent to \begin{gather} \forall y\in Y,((y\in \overline{C})\Rightarrow(x\notin A))\tag2\\ \forall y\in Y,((y\in \overline{B})\Rightarrow(x\notin A))\tag3 \end{gather} which can be combined into \begin{gather} \forall y\in Y,\left(\left[y\in \left( \overline{B} \cup \overline{C}\right)\right]\Rightarrow(x\notin A)\right).\tag{2 and 3}\\ \end{gather} and De Morgan's law gives \begin{gather} \forall y\in Y,\left(\left[y\notin {\left( B \cap C\right)}\right]\Rightarrow(x\notin A)\right)\tag{2 and 3}\\ \end{gather}

From statement (1), we can apply the rule that $\lnot\exists$ is equivalent to $\forall\lnot$: \begin{gather} \left[\forall y\in Y, y\notin (B\cap C)\right]\Rightarrow(x\notin A).\tag1 \end{gather}

Compare the last two statements, which are versions of (1) and (2 and 3). Note that they are similar but not the same, because the parentheses are in different places. Statement (1) says "if $B$ and $C$ are disjoint, then $x\notin A$". Statement (2 and 3) says "for all $y$, if $y$ is outside the intersection of $B$ and $C$, then $x\notin A$".

Therefore, (1) is not equivalent to (2 and 3). The following is a counterexample: $$x=1,\qquad A=\{1\}, \qquad B=\{1,2\}, \qquad C=\{2,3\}$$ With these values,

• your original statement evaluates to $\textrm{true} \Rightarrow \textrm{true}$, which is true
• statement (1) evaluates to $\textrm{false} \Rightarrow \textrm{false}$, which is true
• statement (2 and 3) is false, because for $y=3$, we get $\textrm{true} \Rightarrow \textrm{false}$, which is false
• statement (2) is true for $y=3$
• statement (3) is false for $y=3$.