Prove $P(\emptyset)=\{\emptyset\}$ where $P(\emptyset)$ is the power set of the empty set.

My doubt is whether or not it's advisable to harness a lemma in this proof so that it reads more smoothly. The aforesaid lemma would read:

Lemma: $\exists!\, S:S\subseteq \emptyset$

Proof. By virtue our lemma, indeed$$\forall x:\big(x\subseteq \emptyset \iff x=\emptyset\big)\iff \forall x:\big(x\in P(\emptyset)\iff x\in \{\emptyset\}\big)\iff P(\emptyset)=\{\emptyset\}$$

In light of your latest comment, I'll explain why I think the result's proof should be presented without stating the proposed lemma.

Since the desired theorem may be stated as $\forall x(x\subseteq\emptyset\iff x=\emptyset)$, the stated lemma is the crux of the problem, but we can argue as follows directly (as I think you probably already know):

The $\Longleftarrow$ direction is trivial; for the $\implies$ direction, note if $x\subseteq\emptyset$ then there are no elements of $x$ (as they'd also be elements of $\emptyset$), so $x=\emptyset$ by extensionality.

That looks simple enough to me. If anything, I'd rather regard the lemma as a restatement of this direct result. Indeed, one would usually prove a theorem of the form $\forall x(\varphi(x)\iff x=a)$, then note $\exists!x(\varphi(x))$ as a corollary. (In your case, the corollary makes obvious the identity of $a$ - because its choice of $\varphi$ is the second-argument-fixed case of an unsaturated ordering relation - but that doesn't change my recommendation.) How else would one get to the latter, especially with pedagogical clarity?