Does considering a pair of vacuously true statements lead to a contradiction?

Question

If $A$ is false, then both $A \Rightarrow B$ as well as $A \Rightarrow \lnot B$ are true.

Consider any $X \subset M$.

$ \forall \alpha, \alpha \in \emptyset \Rightarrow \alpha \in X$ is true, since $ \alpha \in \emptyset $ is always false.

Therefore, $\emptyset \subset X$ $......(1) $

$ \exists \alpha, \alpha \in \emptyset \Rightarrow \alpha \notin X$ is also vacuously true for the same reason.

Thus, $\emptyset \not \subset X$ $..........(2) $

$(1)$ and $(2)$ lead to a contradiction.

Answer 1

The second statement is not entirely correct.

The relation $A \nsubseteq B$ means that $$(\exists x) \big( x\in A \, \wedge \, x\notin B \big)$$ but it doesn't mean that $$(\exists x) \big( x\in A \, \Rightarrow \, x\notin B \big)$$

Answer 2

How do you conclude that "Thus, $\varnothing\not\subset X$"? By definition $$\varnothing\subset X\qquad\Leftrightarrow\qquad(\forall x)(x\in\varnothing\ \ \Rightarrow\ \ x\in X),$$ where the latter is vacuously true, as you already note yourself.