Is it possible to prove reflexive, symmetric and transitive properties of equality and the transitive property of inequality? This may be a bit of a trivial question, but can one prove the reflexive, symmetric and transitive properties of equality and the transitive property of inequality of real numbers?(and if so, how? Is there a fairly straightforward, possibly algebraic method?) i.e. Is it possible to prove that $\forall\;a, b\in\mathbb{R}$
$$a=a,$$ $$\text{if } a=b \text{ then } b=a,$$  and $$\text{if } a=b \text{ and } b=c \text{ then } a=c.$$  Also, $$\text{if } a<b \text{ and } b<c \text{ then } a<b$$
The sources that I've read so far seem to have conflicting points of view. 

Absolutely. The equality relation on the real line is stated formally as follows: 
$$S\subseteq R^2 = \{(x,x)|x\in R\}$$ 
Naturally,we assume $S\neq \emptyset$.So let's check all the axioms for an equivalence relation. 
(1) Reflexivity. Clearly for every $x \in R$ , $(x,x)\in S$. 
(2) Symmetry: Let a = b where $a,b\in R$. Then $(a,b)\in S$. 2 ordered pairs in a relation S are the same iff for $(a,b) ,(c,d) \in S$,then a=c and b=d i.e. (a,b) = {{a},{a,b}} = {{c},{c,d}}=(c,d). So since a=b, {{a},{a,b}} = {{b},{b,a}}. But this means $(b,a) \in S$ and b=a.  
(3) Transitivity: Let a=b and b=c where $a,b,c\in R$. That means $(a,b), (b,c) \in S$. By reflexivity, b=b. Since a=b, (b,c) = (a,c). So $(a,c) \in S$. Since $(b,c)\in S$, $(c,b)\in S$ by symmetry. Since a=b, $(c,a)\in S$. But now, since $(a,c) and (c,a)\in S$, then a=c and that does it. So equality on R is an equivalence relation.      
For inequality, a stricter ordering relation then "=" is needed. You have the right idea with your proof,but you have to be a little more careful about the axioms and make sure the order relation is defined via ordered pairs as we've done above. You've got the right idea,though. See if you can finish it yourself. 
