Why is my $\epsilon$-$\delta$ proof incorrect? Prove that $f(x)= \sqrt{4+x^2}$ is continuous at $x_0$ using the $\epsilon -\delta$ definition of continuity.
Textbook proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\underbrace{\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}}_{K(x,x_0)}|x-x_0|$
$K(x,x_0)=\frac{|x+x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq \frac{|x|+|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}+\frac{|x_0|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}$
$\leq\frac{|x|}{\sqrt{4+x^2}}+\frac{|x_0|}{\sqrt{4+{x_0}^2}}$
$\frac{|x|}{\sqrt{4+x^2}}\leq 1$ and $\frac{|x_0|}{\sqrt{4+{x_0}^2}}\leq 1\Rightarrow K(x,x_0)\leq 2\Rightarrow |f(x)-f(x_0)|\leq2|x-x_0|<2\delta=\epsilon$
So, for all $\epsilon$ we can choose a $\delta$ such that $\delta=\frac{\epsilon}{2}$.
My proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}\leq |x^2-{x_0}^2|$
$=|x-x_0||x-x_0+2x_0|<\delta^2+2|x_0|\delta$
Case $\delta<1$:
$\delta^2+2|x_0|\delta<\delta(1+2|x_0|)=\epsilon$
$\Rightarrow$ If $1+2|x_0|>\epsilon$ we have $\delta=\frac{\epsilon}{1+2|x_0|}$
Case $\delta=1$:
$\Rightarrow$ If $1+2|x_0|\leq\epsilon$ we have $\delta = 1$
Why do I get a different result for delta? What did I do wrong?

Regarding your proof, there is nothing wrong with it. Only that dealing with $\delta=1$ is not needed, as you can always request $\delta<1$. 
As for why the "results" are different? There is no result. No one says that for a given $f$ and a given $x_0$ and a given $\varepsilon$, there is a single $\delta$. Any $\delta$ less than a given $\delta$ will work, so there is not obvious choice of a "best" $\delta$ or anything like that. 
