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 looks like you are assuming your claim: you are showing that your claim implies your assumptions $0 < a < b$, which means you should reverse the argument. Try to do so. Here below is a different approach.
Assuming $0 < a < b$ you have $2\sqrt{a(b-a)}>0$, hence
$$2\sqrt{a(b-a)}+a+(b-a) > b.$$
The left hand side is $(\sqrt{b-a}+\sqrt{a})^2$, so 
$$(\sqrt{b-a}+\sqrt{a})^2 > \sqrt{b}^2$$
and since both sides are positive you can take the square root getting 
$$\sqrt{b-a}+\sqrt{a} > \sqrt{b},$$
which is your claim.
