Proof that the empty set is a relation

The reasoning is valid up to the last step. It is true that the empty set is also a set of singletons. But why can't a set of singletons also be a relation? If you think about how you would prove this, you would need to take an element of the set (which is a singleton) and show that element is not an ordered pair. But in this case, there are no elements to take!

(Similarly, the empty set is also a set of real numbers, and a set of cows, and a set of kings of Canada. This doesn't mean that there is a cow that is king of Canada!)

There’s no actual contradiction here. Let $A$ be any set, and let $S=\big\{\{a\}:a\in A\big\}$, the set of singletons of elements of $A$. Then $\varnothing\subseteq S$, so $\varnothing$ can be described (somewhat confusingly) as a set of singletons, but this is so vacuously: it’s a set of singletons because it does not contain anything that isn’t a singleton, not because it actually contains any singletons. Similarly, $\varnothing$ is a set of ordered pairs of elements of $A$, but only vacuously so, in that it does not contain anything that isn’t such an ordered pair. In fact, if $X$ is any set, $\varnothing$ could be called a set of elements of $X$, simply because $\varnothing\subseteq X$, but this is only vacuously true. It’s best just to notice that $\varnothing$ is a subset of every set and not to try to talk about the nature of its (non-existent) elements.

In particular, it’s better to say simply that every subset of $A\times A$ is by definition a relation on $A$ and then note that $\varnothing\subseteq A\times A$.

The empty set is a set of singletons and a set of pairs and a set of topological spaces and a set all of whose members are unicorns. That may seem contradictory, but it isn't. The reason for this lies in the underlying logic. Let $U(x)$ be a predicate that is true of $x$ iff $x$ is a unicorn and consider the sentence $$ \forall x \colon x \in \emptyset \to U(x). $$ This sentence is true, because "$x \in \emptyset$" is false for any $x$. So the right hand side actually doesn't matter and for any formula $\varphi(x)$ the sentence $$ \forall x \colon x \in \emptyset \to \varphi(x) $$ is true by the same reasoning.

So where exactly does your attempt to prove that $\emptyset$ is no relation go wrong? Well... "$x$ is a relation" may be formalized in the following way: Let $\psi(x)$ be the formula $$ \forall y \in x \exists a \exists b \colon y = (a,b) $$ (Note that "$y = (a,b)$" is an abbreviation for a first order formula $\pi(y,a,b)$ that is satisfied iff $y = (a,b)$.)

If $\emptyset$ were not a relation, then $\neg \psi(\emptyset)$ would be true, i.e. $$ \exists y \in \emptyset \forall a \forall b \colon y \neq (a,b) $$ or more precisely $$ \exists y \forall a \forall b \colon y \in \emptyset \wedge y \neq (a,b). $$ However, $y \in \emptyset$ is false for any $y$ and thus $\neg \psi(\emptyset)$ is also false. Therefore $\psi(\emptyset)$ must be true and $\emptyset$ is proved to be a relation. By changing $\psi$ in the obvious way, we may also prove that $\emptyset$ is a set of singletons or a set of unicorns or...