Here's my proof but I'm looking for a simpler or at least direct proof for the indirect part of this proof.
Proof. Given two sets, say $a$ and $x$, it all boils down to stablish.\begin{align*}x\subseteq \{a\}\iff x=\emptyset\vee x=\{a\}\end{align*}
($\impliedby$)Trivial.
It's trivial once we know $\forall S: S\subseteq S\wedge \emptyset \subseteq S$
($\implies$)Let us assume now, $$x\subseteq \{a\}\wedge x\ne \emptyset\tag{*}$$If there is $t\in\{a\}: t\notin x$ by the uniqueness of the empty set we know $x=\emptyset$, a contradiction to $(*)$ So it must be that $\forall t:\big(t\in \{a\}\implies t\in x\big)$. Furthermore, we know $\forall t:\big(t\in \{a\}\iff t\in x\big)$ also because of $(*)$Then,\begin{align*}\Big((*)\implies x=\{a\}\Big)&\iff \Big(x\subseteq \{a\}\implies \big(x\notin \emptyset\implies x=\{a\}\big)\Big)\\
&\iff \Big(x\subseteq \{a\}\implies x=\emptyset \vee x=\{a\}\Big)\end{align*}
as we intended to prove.
Best Answer
Similar to the proof by Hermis14 but more direct, avoiding contraposition. As already observed, it suffices to prove that $\forall x\,(x\subseteq\{a\} \implies(x=\varnothing \lor x=\{a\}))$. So consider any $x\subseteq \{a\}$. Thus all members of $x$ (if any) are equal to $a$.
Case 1: $a\in x$. Then $a$ is the one and only member of $x$, so $x=\{a\}$.
Case 2: $a\notin x$. Then $x$ has no members at all, so $x=\varnothing$.
In both cases, $x=\varnothing \lor x=\{a\}$ as required.