Proof of intersection and union of Set A with Empty Set

This looks fine, but you could point out a few more details. For instance, $x\in \varnothing$ is always false. Therefore $x \in A \text{ or } x\in \varnothing $ is logically equivalent to $ x \in A $ and therefore the two set descriptions $$ \{x \mid x \in A \text{ or } x \in \varnothing\},\quad \{x\mid x \in A\} $$ must describe the same set, since the conditions are true for exactly the same elements $x$.

Similarily, because $x \in \varnothing$ is trivially false, the condition $x \in A \text{ and } x \in \varnothing$ will always be false, so the two set descriptions $$ \{x \mid x \in A \text{ and } x \in \varnothing\},\quad \{x\mid x \in \varnothing \} $$ must describe the same set.

Of course, for any set $B$ we have $$ B = \{x \mid x \in B\} $$

Yes. The solution works, although I'd express the second last step slightly differently.

$\begin{align} A\cup \varnothing & = \{x:x\in A \vee x\in\varnothing \} & \text{definition of union} \\ &= \{x:x\in A \} & \neg\exists x~(x\in \varnothing) \\ & = A \\[2ex] A\cap\varnothing & = \{x:x\in A \wedge x\in \varnothing \} & \text{definition of intersection} \\ & = \{\} & \neg\exists x~(x\in \varnothing \wedge x\in A) \\ & = \varnothing \end{align}$

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$A\cup \varnothing = A$ because, as there are no elements in the empty set to include in the union therefore all the elements in $A$ are all the elements in the union. Hence the union of any set with an empty set is the set.

$A\cap \varnothing = \varnothing$ because, as there are no elements in the empty set, none of the elements in $A$ are also in the empty set, so the intersection is empty. Hence the intersection of any set and an empty set is an empty set.