Let $U = \{s_1, s_2, s_\cdots, s_n\}$ be the universal set and $A = \{a_1, a_2, a_3,\cdots, a_m\}$ is the set under consideration.
Now, I am proving $\forall_{x \in \{\}} p(x) = T$ as follows
$$\forall_{x \in A} p(x) = \forall_{x \in u} (x \in A \implies p(x)) $$ $$= (s_1 \in A \implies p(s_1)) \land (s_2 \in A \implies p(s_2)) \land \cdots (s_n \in A \implies p(s_n))$$
Hence if $A = \{\}$, then premise in every conditional statement becomes false and the result will be true.
But it is not working to prove $\exists_{x \in \{\}} p(x) = T$;
How to prove it?
This won't work because $\exists x\in\{\},p(x)$ is false.
Like Mauro says in the first comment, $ \exists x\in\{\},p(x) $ is syntactic sugar for $\exists x , (x\in\{\} \wedge p(x))$. But $x\in\{\}$ can never be true, so $x\in\{\} \wedge p(x)$ is never true, so $\exists x , (x\in\{\} \wedge p(x))$ is false.
You seem hung up on expanding quantification to the individual elements of a set. That's not necessary, and seems to limit you logically to a finite universal set. But these statements are true irrespective of the cardinality of $U$.