I would say that a relation should not be called "reflexive", without further qualification, unless the underlying set which the relation is on is specified (explicitly or implicitly).
For example your set of ordered pairs $R$ is a subset of $\mathbb{R} \times \mathbb{R}$ and hence is a relation on $\mathbb{R}$. It is not reflexive on $\mathbb{R}$ because there are real numbers $x$ with the property that $(x,x)$ is not in $\mathbb{R}$. But your set $R$ is also a subset of $(0, 10) \times (0,10)$ and hence is a relation on the open interval $I = (0,10)$, and it is is a reflexive relation on this set. To abstract this somewhat, the lesson is that whether a set of ordered pairs is reflexive or not, depends on more than just the set of ordered pairs. It depends on the set $A$ that you think of the set of ordered pairs as being a subset of, and this is not encoded in the set $R$ at all.
(This is a little bit like how the codomain of a function is not determined by the set of ordered pairs that represents the graph of a function.)
As you have defined them the notions of "symmetric" and "transitive" also depend on the underlying set, but this dependence is less sensitive, in the following sense: if $S$ is a set and $R$ is a subset of $S \times S$, and $R$ is symmetric (or transitive) as a relation on $S$, then for any set $T$ satisfying $T \supseteq S$ (so you have that $R$ is also a subset of $T \times T$, and hence a relation on $T$), $R$ will also be symmetric (or transitive) as a relation on $T$. As the above example shows, reflexivity does not behave like this. So symmetry and transitivity are far less sensitive to what set you consider a set of ordered pairs to be relation on, than reflexivity is. So you should not be casual about saying that a relation is "reflexive" (without further qualification) unless a reader knows from context what set you have in mind. (In actual mathematical practice this is never an issue because relations are not constructed arbitrarily for the purpose of testing knowledge of concepts like "reflexivity". But in classroom exercises it is a good practice to follow.)
If I may veer into metamathematics for a moment. Speaking very informally, reflexivity is fundamentally different in nature from the symmetry and transitivity because it is the only property that actually requires specific ordered pairs to be in a relation. (The other two only require that if some ordered pairs are in there, then other ordered pairs must be in there also.) In textbook exercises often people think of reflexivity as a silly and trivial property because it is easy to verify in many examples whether or not a relation is reflexive (on a given set). This is more an artifact of what happens in textbooks than it is an artifact of actual mathematical practice. In actual mathematical practice, it is often the case that reflexivity (when it occurs) is the most difficult property of these three to verify--- precisely because it is the only one that requires specific ordered pairs to be in a relation, it is the only one where in proving it you might be required to actually do something with the definition of the relation and not just use its formal properties.
As an illustration of this idea, suppose I let $A$ denote the set of simple closed curves in $\mathbb{R}^3$ and say that $C \sim D$ for $C, D \in A$ iff $C$ there is a surface $S$ in $\mathbb{R}^3$ satisfying [a long list of geometrical constraints] with the property that the boundary of $S$ is the disjoint union of a curve homeomorphic to $C$ and another curve homeomorphic to $D$. In checking that $\sim$ is symmetric and transitive, I only need to think about what happens to my long list of geometrical constraints if I permute the order of the data I feed into it, or if I piece two surfaces satisfying these constraints together. To prove that $\sim$ is reflexive I actually have to construct, for every curve $C$, a surface satisfying [long list of constraints] whose boundary is a disjoint union of two copies of $C$. A totally different kind of problem.
If neither $A$ nor $\neg A$ is a theorem of $T$, then both $T\cup\{\neg A\}$ and $T\cup\{A\}$ must be satisfiable - by the Lowenheim-Skolem theorem, that means that there are denumerable models.
Now when we add (say) $A$ to $T$, we "lose" a(n isomorphism type of a) denumerable model - namely, the model of $T\cup\{\neg A\}$. Since $T$ only had two denumerable models to begin with, how many does that leave?
Best Answer
I'd say the simplest decidable-but-incomplete theory is the (set of consequences of the) single axiom "$\exists x,y\forall z(x=z\vee y=z)$" (in the empty language). This sentence has exactly two models up to isomorphism, the one-element structure and the two-element structure, and each of these structures is decidable; any theory with finitely many (up to isomorphism) at-most-countable models, each of which is decidable, is itself decidable. A model is decidable if its theory is decidable; note that the theory of a model is always a complete theory.
A less silly trick is the following: given sets $\Gamma,\Delta$ of sentences, consider the new set $$\Gamma\oplus\Delta=\{\gamma\vee\delta: \gamma\in\Gamma,\delta\in\Delta\}.$$ It's easy to check that every model of $\Gamma$ is a model of $\Gamma\oplus\Delta$, and every model of $\Delta$ is a model of $\Gamma\oplus\Delta$. The nontrivial observation is that the converse is also true! Suppose $\mathcal{M}\not\models\Gamma$ and $\mathcal{M}\not\models\Delta$. Then we can find $\gamma\in\Gamma$ and $\delta\in\Delta$ such that $\mathcal{M}\not\models\gamma$ and $\mathcal{M}\not\models\delta$. But then $\mathcal{M}\not\models(\gamma\vee\delta)$, and so $\mathcal{M}\not\models\Gamma\oplus\Delta$.
So the class of models of $\Gamma\oplus\Delta$ is exactly the union of the class of models of $\Gamma$ and the class of models of $\Delta$. In particular, if $\Gamma$ and $\Delta$ are each decidable, then so is $\Gamma\oplus\Delta$: we have $\Gamma\oplus\Delta\vdash\varphi$ iff $\Gamma\vdash\varphi$ and $\Delta\vdash\varphi$. So this lets us combine decidable complete theories to get decidable incomplete theories.