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.

You just need to note that every step you did is reversible. For the conclusion, from $a\le\sqrt{ab}$ you get the equivalent $a^2<ab$ and, dividing by $a$, $a<b$, which is true.
More simply, set $x=\sqrt{a}$ and $y=\sqrt{b}$. You need to show that
$$
y-x<\sqrt{y^2-x^2}
$$
Since $y>x$, this is equivalent to
$$
(y-x)^2<y^2-x^2
$$
that is,
$$
y^2-2xy+x^2<y^2-x^2
$$
which in turn is equivalent, with simple algebraic simplifications, to
$$
x^2<xy
$$
which is true because $x<y$ and so $x^2<xy$.
