Equalizers don't exist in Rel. I suppose that there are equalisers in the category of relations Rel.
It contradicts to https://ncatlab.org/nlab/show/Rel :
"... order-relation $'\leq' \; \subseteq \{0,1\} \times \{0,1\}$ can’t be split. It follows that it can’t have (co)equalisers."
I agree that leq is not split.
But empty relation between every two objects always exists!
So (especially when one or both of relations $f$ and $g$ are not functions) then every set(which is object in Rel) is an equaliser.
Empty set is always apex of one of the cones.
 "Empty relation" $ = \emptyset \subset \emptyset = A \times \emptyset =$ "set of relations between $A$ and empty set"
But, of course, not all sets are isomorphic. So my statement cannot be true. Where is my mistake?  
I used definition from wikipedia:
"The equaliser consists of an object $E$ and a morphism $eq : E → X$ satisfying $f\circ eq=g\circ eq$, and such that, given any object $O$ and morphism $m : O → X$, if $f\circ m=g\circ m$, then there exists a unique morphism $u : O → E$ such that $eq\circ u=m$."
 A: You're misreading; by saying "A category $\mathcal{C}$ doesn't have [(co)limits of a given type]", what one means is that there exists a diagram of the given type that fails to have a (co)limit.
That is, such a statement is the negation of the assertion that every diagram of the given type has a (co)limit.
For many types of diagrams, you are always guaranteed that a nonempty category has some instances of the diagram that have (co)imits; e.g. the constant diagram (all objects the same, all morphisms identities) often works.

From your comments, I think you're claiming that, in Rel, the empty set is an equalizer of the two morphisms $\leq$ and $1_{\{0,1\}}$.
Letting $e$ be the empty relation $\{\} \to \{0,1\}$, it is indeed true that $\leq \circ e = 1_{0,1} \circ e$. But that does not imply that $e$ is the equalizer!
To wit, $\leq \circ \leq = 1_{0,1} \circ \leq$, so if $e$ was the equalizer, there would exist a unique morphism $p : \{0,1\} \to \{ \}$ such that $\leq = e \circ p$.
However, the only morphism $p : \{0,1\} \to \{\}$ is the empty relation, and $e \circ p$ is not the relation $\leq$.
