Proof Verification: Spivak's Calculus, Chapter 1, Problem 8 This is a proof for Spivak's Calculus 4th ed., Chapter 1, Problem 8.

Prove that for every number $a$, one and only one of the
  following holds:
(i) $a = 0$,
(ii) $a$ is in the collection $P$ (all the positive numbers), 
(iii) $—a$ is in the collection $P$.
(The Trichotomy Law)

Here are the properties that I was allowed to use

$(PI)$ (Associative law for addition) $a + (b + c) = (a + b) + c$
$(P2)$ (Existence of an additive identity) $a+0=0+a=a$
$(P3)$ (Existence of additive inverses) $a+(-a)=(-a)+a=0$
$(P4)$ (Commutative law for addition) $a+b=b+a$
$(P5)$ (Associative law for multiplication) $a\cdot(b\cdot c)=(a\cdot b)\cdot c$
$(P6)$ (Existence of a multiplicative identity) $a\cdot 1=1\cdot a=a, 1\neq0$
$(P7)$ (Existence of multiplicative inverses) $a\cdot a^{-1}=a^{-1}\cdot a = 1$, for $a\neq0$
$(P8)$ (Commutative law for multiplication) $a\cdot b = b\cdot a$
$(P9)$ (Distributive law) $a(b+c)= ab+ac$
$(P'10)$ For any numbers $a$ and $b$, one and only one holds:
(i) $a=b$
(ii) $a>b$
(iii) $a<b$
$(P'11)$ For any numbers $a$, $b$ and $c$, if $a<b$ and $b<c$, then $a<c$.
$(P'12)$ For any numbers $a$, $b$ and $c$, if $a<b$, then $a+c<b+c$.
$(P'13)$ For any numbers $a$, $b$ and $c$, if $a<b$ and $c>0$, then $ac<bc$.

And here is my proof:
If, for two numbers $a$ and $b$, $a=b$, it is easy to prove that $b-a=0$.
If, $a<b$, it is easy to prove that $0<b-a$.
If, $a>b$, it is easy to prove that $0<-(b-a)$.
$\therefore$ As any real number can be expressed as the sum of two other real numbers, I can let $x=b-a$ such that either $x=0$ or $x>0$ or $-x>0$.
My worry isn't proving the "this is easy to prove" parts as I'm fine with those. However, I am concerned if I can assume that I can let any number $x=b-a$. 
Does this proof make sense?
 A: You can certainly give a name ($x$) to the difference of two real numbers ($b$ and $a$).  That follows from (P3), which essentially says that the real numbers are closed under subtraction.
But I'm a bit confused by your use of $b$ in the first place.  The statement you are asked to prove is:

For all real numbers $a$, either $a \in P$, $-a \in P$, or $a=0$.

There's no reference to a second number $b$, but your proof creates one and proves statements about $b-a$.
However, you can use a specific number $b$ and invoke (P'10) to prove the statements required.
A: "I am concerned if I can assume that I can let any number x=b−a"
Well, yes and no but mostly no.
You were asked to prove:  For any number, $symbol$, that something is true for $symbol$.
You proved: For any two numbers, $a,b$ something is true for $a-b$.  If we let $symbol = a-b$ then it is true for $symbol$.  So something is true for all numbers $symbol$ where $symbol$ may be written as a difference of two numbers.
But you have neither considered numbers that can't be written as a difference of two numbers, nor have you considered proving that any number $symbol$ can be written as a difference of two numbers.
If you do either of those the proof is complete.
Proving that any number $symbol$ is easy:  Pick any number $d$ at all (some numbers do exist so this is possible).  Let $a = symbol + d$ and $b = d$ then $a - d = symbol + d - d = symbol$ and .... ta-da.
But really your proof can be much easy:
P'(10) says that for any two numbers $a$ and $b$ that either $a < b; a=b;$  or $a < 0$.  Let $a$ be any number and $b$ be $0$.  Then either $a < 0$ or $a = 0$ or $a > 0$.
I assume that when you said it "was easy" to prove that if $a > b$ then $b-a < 0$ that it will be just as easy to prove if $a < 0$ that $-a > 0$.
