Is this $\delta-\varepsilon$ proof correct?

Let $\Omega$ be an intervall and let $\gamma\in\Omega$. $\lim_{x\rightarrow \gamma}f(x)=L$ is equivalent to the statement bellow.

$$\forall \varepsilon>0\exists \delta:\forall x \in \Omega: 0<\mid x-\gamma\mid<\delta \Rightarrow \mid f(x)-L \mid < \varepsilon$$

I'm working on an assignment where I have to use the above definition of a $\lim_{x\rightarrow \gamma}f(x)=L$ to show that a particular function converges. I've come up with two proofs thus far but I am not certain that the second one is correct since I have allowed $\Omega$ to depend on $\varepsilon$. Is this allowed?

I will give you an example of what I did. At a certain point in the proof I find that $\mid f(x)-L \mid = C\mid x-\gamma\mid ^2$, $C>0$. So if $0<\mid x-\gamma\mid<\sqrt{\frac{\varepsilon}{C}}$ we have that $\mid f(x)-L \mid = C\mid x-\gamma\mid ^2 < C\left(\sqrt{\frac{\varepsilon}{C}}\right)^2=\varepsilon$. If I let $\Omega = \left(\gamma - \sqrt{\frac{\varepsilon}{C}}, \gamma + \sqrt{\frac{\varepsilon}{C}}\right)$ then I can let $\delta = \sqrt{\frac{\varepsilon}{C}}$ correct? Am I allowed to do this or did I mess up?

No you cannot change $\Omega$; but there is no need to anyway. You've already solved it. You can choose $\delta = \sqrt{\frac{\varepsilon}{C}}$ regardless of what $\Omega$ is. It doesn't matter if your $\Omega$ is contained in $(\gamma - \delta, \gamma + \delta)$ or if $(\gamma - \delta, \gamma + \delta)$ is contained in $\Omega$, or neither. To prove the limit is $L$ you must prove that for every $x$ which is in $\Omega$ and in $(\gamma - \delta, \gamma + \delta)$, we have $|f(x)-L|< \epsilon$. It doesn't matter if some points are in $\Omega$ but not in $(\gamma - \delta, \gamma + \delta)$ or vice versa.