Let $A$ be any commutative ring and consider $\operatorname{Spec}(A)$, then for any $E \subseteq A$ we have $V(E) = \emptyset \iff E = A$
Recall that $V(E) = \{\text{prime ideals of $A$ containing $E$}\}$
Proof: That Since $V(A) = V(1) = \emptyset$ the reverse direction is trivial. To prove the forward direction suppose that $V(E) = \emptyset$, then no prime ideal of $A$ contains $E$. Let $(E)$ be the ideal generated by $E$, then we have $V(E) = V(\operatorname{rad}(E)) = \emptyset$ and so no prime ideal contains the intersection of all prime ideals containing $E$.
Now if $(E) \neq (1)$ then $(E) \subsetneq M$ for some maximal ideal of $A$ containing $(E)$. But then $M$ is a prime ideal containing $(E)$ and so must contain the intersection of all prime ideals containing $E$, so $M \in V(E)$ a contradiction. Thus $(E) = (1) = A$. $\square$
I just wanted to check if my proof was correct and my reasoning was sound and efficient.
For starters, surely $E$ itself is supposed to be an ideal and not just a subset (or even subring). Otherwise consider $E = \{1 \}$ for which (generally) $E \not= A$ but $V(E) = \emptyset$. Also, when you say $E \subsetneq M$, that should be $E \subseteq M$, since $E$ might itself be a prime ideal.
Beyond that, your proof is correct but needlessly complicated. Your reasoning is perfectly fine without mentioning radicals and intersections of primes. You also are structuring your reasoning like a proof by contradiction but proceeding by contraposition, which is confusing.
Indeed, what you are really saying is: $$E \not= A \implies 1 \notin E \implies E \subseteq M \text{ for some maximal ideal $M$ by Zorn's lemma} \implies V(E) \not= \emptyset \text{ by definition}$$
Contrapositively, $V(E) = \emptyset \implies E = A$