If x and y are distinct real numbers prove there's a neighborhood P of x and neighborhood Q of Y such that $P\cap Q = \emptyset$ So here is the proof that I have: 
By definition, $\exists \varepsilon > 0$ so that $(x-\varepsilon, x+\varepsilon) \subset P$ and $(y-\varepsilon, y+\varepsilon) \subset Q$
Choose $\varepsilon = \frac{|x-y|}{2} > 0$ 
Assume that $z \in P\cap Q$
That implies that $|x-z| < \varepsilon$ and $|z-y| < \varepsilon$ 
$\Rightarrow |x-z| + |z-y| < 2\varepsilon = |x-y|$
Since $z \in P\cap Q$ than z must lie between x and y 
$\Rightarrow |x-y| \leq |x-z| + |z-y| < |x-y|$
$\Rightarrow \Leftarrow$ Since $|x-z| + |z-y|$ cannot be strictly less than $|x-y|$ and also greater than or equal to $|x-y|$ 
Therefore we have that $P\cap Q = \emptyset$
 A: You’ve the right understanding of how to get disjoint nbhds of $x$ and $y$, but your proof is very badly organized.

By definition, $\exists \varepsilon > 0$ so that $(x-\varepsilon, x+\varepsilon) \subset P$ and $(y-\varepsilon, y+\varepsilon) \subset Q$

How can there be anything involving $P$ and $Q$ at this point? You haven’t yet constructed $P$ and $Q$, so the names ‘$P$’ and ‘$Q$’ don’t actually signify anything yet. I would drop this altogether and start with your next sentence:

Choose $\varepsilon = \frac{|x-y|}{2} > 0$,

though I’d say

Let $\epsilon=\dfrac{|x-y|}2>0$.

Quoting again:

Assume that $z \in P\cap Q$

Like the first sentence, this is meaningless: as yet we have no $P$ or $Q$. Before you can say this, you need to define $P$ and $Q$:

Let $P=(x-\epsilon,x+\epsilon)$ and $Q=(y-\epsilon,y+\epsilon)$. Now suppose that $z\in P\cap Q$.

Quoting again:

That implies that $|x-z| < \varepsilon$ and $|z-y| < \varepsilon$
$\Rightarrow |x-z| + |z-y| < 2\varepsilon = |x-y|$

I’d use words rather than $\Rightarrow$:
That’s fine mathematically, though it reads more smoothly if you use words:

This implies that $|x-z|<\varepsilon$ and $|z-y|<\varepsilon$ and hence that $|x-z|+|z-y|<2\epsilon=|x-y|$.

In any case it wants a period/full stop at the end, as do your other sentences. Quoting again:

Since $z \in P\cap Q$ than z must lie between x and y

While it’s true that $z$ must lie between $x$ and $y$, it’s also irrelevant. What you really want here is simply the triangle inequality. In fact, you can combine the final step with the previous one:

This implies that $|x-z|<\varepsilon$ and $|z-y|<\varepsilon$ and hence that $$|x-y|\le|x-z|+|z-y|<2\epsilon=|x-y|\;,$$ which is an obvious contradiction.

Quoting again (after omitting a couple of unnecessary lines):

Therefore we have that $P\cap Q = \emptyset$

A: The sentence "since $z\in P\cap Q$ than $z$ must lie between $x$ and $y$" is wrong. At this point you just know that $z$ is close to both $x$ and $y$; so close that it will turn out to be impossible. But before you have proved that you may not assume anything about $z$, in particular not that it is between $x$ and $y$, even if that is intuitively the best shot at the (impossible) task of being less that $\epsilon$ away from both $x$ and $y$. So you must do without that assumption about$~z$.
