There is relation that is symmetric and transitive but not reflexive? Let $L=\left\{R\right\}$ be a language with only one relation symbol.
Consider these formulas:
$\Psi _1\:=\:\forall x\left(R\left(x,x\right)\right)$
$\Psi _2\:=\:\forall x\forall y\left(R\left(x,y\right)\rightarrow R\left(y,x\right)\right)\:$
$\Psi _3\:=\:\forall x\forall y\forall z\left(R\left(x,y\right)\wedge R\left(y,z\right)\rightarrow R\left(x,z\right)\right)\:$
I need to show $Mod\left\{\Psi _1,\Psi _2,\Psi _3\right\}\ne Mod\left\{\Psi \:_2,\Psi \:_3\right\}$.  

So far, i understand the meaning of each formula, but i wonder which relation on any domain, can show that  $R$ is symmetric and transitive but not reflexive. Any ideas?  

EDITED: Well, i looked at If a relation is symmetric and transitive, will it be reflexive? and it really helpfull! but i want to find other examples just from curiosity ..
The empty relation on a nonempty domain is vacuously symmetric and transitive, but not reflexive.
Quite generally, if the relation $R(x,y)$ holds, then from symmetry you have $R(y,x)$ and then from transitivity you have $R(x,x)$. So on the face of it symmetry+transitivity indeed imply reflexivity. However, note that we implicitly assumed that there exists such $y$. 

So to conclude your question: 


  Theorem: For a relation that's symmetric and transitive, and that for every $x$
  there's a $y$ such that $R(x,y)$, then indeed $R$ is also
  reflexive.