Spectrum of Ring is Nonempty

Theorem
$\newcommand{\Spec}{\operatorname{Spec}}$Let $A$ be a non-trivial, commutative ring with unity.

Let $\Spec(A)$ be the prime spectrum of $A$.

Let $\operatorname{Max}\:\operatorname{Spec}(A)$ be the maximal spectrum of $A$.

Then $\Spec(A) \neq \emptyset$ and $\operatorname{Max}\:\operatorname{Spec}(A) \neq \emptyset$

Proof
Let $(\Sigma, \subseteq)$ be the set of all ideals $I \neq A$ under inclusion.

Let $I,J,K \in \Sigma$.

Then it is trivial from the definitions that Therefore $\subseteq$ is an order on $\Sigma$.
 * $I \subseteq I$
 * If $I \subseteq J$ then $J \subseteq I$
 * $I\subseteq J$ and $J \subseteq K$ then $I \subseteq K$.

Let $\{ I_\alpha \}$ be a chain in $\Sigma$.

Claim: $J = \bigcup_\alpha I_\alpha$ is a proper ideal of $A$.

Proof: Indeed, if $J = A$, then $1 \in J$, so for some $\alpha$, $1 \in I_\alpha$.

This implies that $I_\alpha = A$, which contradicts the definition of $\Sigma$.

Now let $a \in A$, $x,y \in J$ be arbitrary.

Then there exist $\alpha$, $\beta$ such that $x \in I_\alpha$, $y \in I_\beta$.

By the definition of a chain, either $I_\alpha \subseteq I_\beta$, or $I_\beta \subseteq I_\alpha$.

Let $K$ be the larger of these two sets, so $x,y \in K$.

Then $K$ is an ideal, so $xy, x \pm y \in K$ and $ax \in K$.

Since $K \subseteq J$ it follows that $J$ is an ideal of $A$.

So we have shown that every non-empty chain in $\Sigma$ has an upper bound in $\Sigma$.

Therefore by Zorn's Lemma, $\Sigma$ has a maximal element.

That is, an ideal $\mathfrak m \neq A$ such that if $I \in \Sigma$ with $\mathfrak m \subseteq I$, then $\mathfrak m = I$.

This is precisely the definition of a maximal ideal.

Therefore, $\operatorname{Max}\:\operatorname{Spec}(A) \neq \emptyset$.

Also because a Maximal Ideal is Prime, $\Spec(A) \neq \emptyset$.