Prove $\sqrt{b} - \sqrt{a} < \sqrt{b-a}$ 
Prove that if $0 < a < b$ then
  $$\sqrt{b} - \sqrt{a} < \sqrt{b-a}$$

This is what I have so far:
square both sides to get $a + b -2\sqrt{ab} < b-a$
subtract $b$ from both sides $a-2\sqrt{ab} < -a$
add $a$ to both sides $2a-2\sqrt{ab} < 0$
than add $2 \sqrt{ab}$ to both sides and get $2a < 2\sqrt{ab}$
divide by $2$ and get $a < \sqrt{ab}$
Thus we know $\sqrt{ab}$ is bigger than $a$ so that $\sqrt{a \cdot a} < \sqrt{ab}$ which means $\sqrt{a} < \sqrt{b}$. 
Therefore together with the given $a < b$, we have $\sqrt{b} - \sqrt{a} < \sqrt{b - a}$
I'm confused as to where I went wrong and how to fix this.

It seems the OPs is somewhat awkwardly presented, and so ascertaining its validity is a bit of a challenge.  If the proof were more concise, it might be easier to decipher.  Here's a suggestion:
We recall that
$0 < a < b; \tag 0$
then, note that
$b = a + (b - a) < a + 2\sqrt a \sqrt{b - a} + (b - a) = (\sqrt a + \sqrt{b - a})^2; \tag 1$
thus,
$\sqrt b < \sqrt a + \sqrt{b - a}, \tag 2$
whence
$\sqrt b - \sqrt a < \sqrt{b - a}. \tag 3$
