Show that $\leq$ on $\mathbb{R}$ satisfies the following three properties. I am new to proofs.
Let $A$ be a relation on $\mathbb{R}$.
Reflexivity: For all $x\in\mathbb{R}, x \leq x$.
Proof: Let $x$ be an element of A. Clearly, $x$ will always equal itself.
This clearly needs more work, but I am having a hard time moving further.
Antisymmetric: For all $x,y\in\mathbb{R}$, if $x\leq y$ and $y\leq x$, then $x=y$.
Proof: Suppose that $x\neq y$. Then, there is some element $x$ and $y$, where $x<y$ or $x>y$. If $x<y$, then clearly $x\not> y$. Similarly, if $y<x$, then clearly $y\not> x$. Thus, $x$ must equal $y$ if $x\leq y$ and $y\leq x$. 
I felt like I did some circular logic here; please help.
Transitivity: I have not attempted this yet. I wanted to make sure I had the right approach on my previous proof.
 A: You cannot let A be just any relation for R
and expect to prove anything much about A.
If you defined A = { (x,y) : x,y in R, x <= y },
then useful propositions about A could be proved.  
To proceed, how are you defining x <= y?
A definition is needed to prove anything about it.  
One definition is:
x <= y when exists nonnegative d with x + d = y.  
Showing reflectivity requires knowing 0 is nonnegative.
Showing transitivity requires knowing the sum of two nonnegative numbers is nonnegative.
Showing antisymmetricity requires knowing that if the
sum of two nonnegative numbers is 0, then they both are.  
How does your text book define <=? 
A: In order to create a proof, you need to have clear definitions.  Without knowing how your course defines $ x \leq y $ for $ x,y \in \mathbb{R} $, it is impossible to generate a satisfactory proof.
Here's a bogus attempt at defining $ x \leq y $ that nonetheless is better than no definition: $ \forall x,y \in \mathbb{R}, x\leq^{*} y \iff x + 1 = y $.  Here, we understand what we mean by $ \leq^{*} $ because we can check whether or not $ x \leq y $ by just converting the problem into one involving addition and equality, both of which I am presuming we have a solid axiomatic understanding.  It turns out that this is the wrong definition because $ -1 \not\leq^{*} 1 $ by the definition above (contrary to what we want), but it is, at least, a definition.  Moreover, this definition fails the reflexive property: $ 0 \neq 0 + 1 = 1 $, so $ 0 \not\leq^{*} 0 $.
But here's a better try: define $ x \leq y $ by $ \exists c \in \mathbb{R}, y = x + c^{2} $.  Now we're onto something!  Below is an attempt at a proof that this definition is good for the properties you've listed.
Reflexive property: $ \forall x \in \mathbb{R}, x = x + 0^{2} $, so $ \forall x \in \mathbb{R}, x \leq x $.
Antisymmetry: if $ x\leq y $ and $ y\leq x$, then $ x = y + c^{2} $ and $ x = y + d^{2} $ for some $ c,d \in \mathbb{R} $.  Then, solve for $ c^{2} $ and $ d^{2} $ and add the equations together to conclude that $ c^{2} + d^{2} = 0 $.  So, $ c = d = 0 $, and so $ x = y $.  
Transitivity: if $ x \leq y $ and $ y \leq z $, then $ y = x + c^{2} $ and $ z = y + d^{2} $.  So, $ z = x + c^{2} + d^{2} $.  If $ c^{2} + d^{2} = \alpha^{2} $, then we can conclude that $ z = x + \alpha^{2} $, and so $ x \leq z $.
However, we've come at a problem!  You cannot show, using this definition, that $ c^{2} + d^{2} = \alpha^{2} $ because that entails showing that $ \sqrt{c^{2} + d^{2}} $ makes sense so that we can define $ \alpha = \sqrt{c^{2} + d^{2}} $.  Somewhat surprisingly, we can't prove this without making some kind of additional assumption lest we fall victim to some kind of weird circular reasoning (since $ \sqrt{x} $ is defined for $ x \geq 0 \iff 0 \leq x $).  However, the Pythagorean theorem should provide adequate justification to assume that this is fine.  (In essence, this definition is good enough to prove the reflexive and antisymmetric properties algebraically, but fails to prove the transitive property without an additional assumption, such as the Pythagorean theorem (and that 3 points in space form a triangle)).
Likely, the definition given in your class is different.  Your proofs will depend upon the definition supplied to you.
