Property 2 of an equivalence relation states that if $a\sim b$ and $b\sim c$ then $a\sim c$.
What is wrong with the following proof that properties 2(symmetry) & 3 (transitivity) imply reflexivity.
Let $a\sim b$, then $b\sim a$, whence by property 3 (using $a=c$), $a\sim a$.
$\varnothing$ is a symmetric and transitive relation on any set. But it is only reflexive as a relation on $\varnothing$ itself.
More generally, being symmetric and transitive are intrinsic properties of the relation, and do not depend what is the set that the relation is "on". Reflexivity depends on an external information, which is an additional set over which the relation is defined.
This means that by enlarging the set over which we define the relation, without adding new ordered pairs, we preserve both symmetry and transitivity, but we will easily destroy reflexivity.