Prove $\forall X [X \subseteq A \land y \in A] \rightarrow \exists X[X \subseteq A \land y \in X]$

Question

Prove: $\forall X [X \subseteq A \land y \in A] \rightarrow \exists X[X \subseteq A \land y \in X]$

proof: Proving by contradiction, suppose

$$\forall X [X \subseteq A \land y \in A] \text{ and } \forall X [ \exists z [z \in X \land z \notin A] \lor y\notin X ]$$

Then because $\forall X [ \exists z [z \in X \land z \notin A] \lor y\notin X ]$, it follows that for ever set $X$, there exists an element say $z$ such that $z \in X$ and $z \notin A$. But this contradicts $\forall X [X \subseteq A \land y \in A]$.

Answer 1

Your answer is true. Other answers:

answer 1.
Your assumption is: $\forall X [X \subseteq A \land y \in A]$ and you want to find an $X$ such that $X \subseteq A \land y \in X$.
By assumption $(A\subseteq A) \land (y \in A)$ (choosing $A$ as one of the every $X$ s in assumption). So the $X$ you look after, can be $A.$

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answer 2. Take the $X$ you look after as $\{y\}$

Answer 2

The premisse is not true: there is no set $A$ that contains every set $X$ as a subset.

So the statement is true.

If $p$ is not true then statement: $$p\implies q$$ or equivalently: $$\neg p\vee q$$ is true.