Suppose $\lim_\limits{x\to0}\sin\frac{1}{x}=A$ where $A\in\mathbb R$.
The negation of the definition of limit is:
$\exists\epsilon>0$ such that $\forall\delta>0$, there is some $x\in\mathbb R$ such that $0<|x|<\delta$ and $\sin\frac{1}{x}\ge\epsilon$.
Let $\epsilon=\frac{1}{2}$. We want to define $x$ so that $|x|$ is less than $\delta$, but $\frac{1}{x}$ is $\frac{\pi}{2}+2\pi n$.
Let $x=\frac{1}{\frac{\pi}{2}+2\pi n}$ where $n$ is a sufficiently large integer so that $|x|<\delta$ (this statement may require more proof, but it is fairly obvious that you can pick such an $n$.
Then $\sin\frac{1}{x}=\sin(\frac{\pi}{2}+2\pi n)=1$.
Voilà. No need for cases.
There are a few issues with your proof.
If $\lim a_n = l_1$ and $\lim a_n = l_2$, the goal is to prove that $l_1 = l_2 = l$.
Assume the opposite: $l_1 \ne l_2$.
You're off to a good start!
$l_1 = l_2$ if $\forall \delta > 0$, $|l_1 - l_2| < \delta$.
$l_1 \ne l_2$ if $\exists \delta > 0$, $|l_1 - l_2| = \delta$.
Both of these statements are true, but neither is directly cogent. The relevant statement is that $l_1\ne l_2$ only if $\exists\delta>0:\left\lvert l_1-l_2\right\rvert\ge\delta.$ This is the contrapositive of the former statement, and is equivalent to the converse of the latter. Even this statement is more formalistic than we really need, though. Instead, we can say something like this: "Let $\delta=\left\lvert l_1-l_2\right\rvert.$ Since $l_1\ne l_2,$ then $\delta>0.$"
(1) If we choose $\epsilon < \dfrac{\delta}{2}$, then $(l_1 - \epsilon, l_1 + \epsilon) \cap (l_2 - \epsilon, l_2 + \epsilon) = \emptyset$.
This is true, but don't forget to require that $\epsilon>0,$ as this statement is true (but useless) even when $\epsilon\le 0.$ Another point worth noting is that we don't actually require $\epsilon<\frac12\delta.$ It is enough to put $\epsilon=\frac12\delta.$ Once you've done that, point out that $\epsilon>0$ since $\delta>0,$ then apply the triangle inequality to prove that $(l_1-\epsilon,l_1+\epsilon)\cap(l_2-\epsilon,l_2+\epsilon)=\emptyset,$ based on our definitions of $\epsilon$ and $\delta$.
(2) If $\lim a_n = l_1$, then by convergence definition $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that if $n > N$, then $|a_n - l_1| < \epsilon$, or $a_n \in (l_1 - \epsilon, l_1 + \epsilon)$.
(3) If $\lim a_n = l_2$, then by convergence definition $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ such that if $n > N$, then $|a_n - l_2| < \epsilon$, or $a_n \in (l_2 - \epsilon, l_2 + \epsilon)$.
Very true.
Choosing such $\epsilon$ that (1) is valid, from (2) it follows: If $\lim a_n = l_1$ then $\lim a_n \ne l_2$.
Be wary of saying "it follows" in a proof, unless you have a previous result to back it up! In fact, it does not follow from (2)--at least, not from (2), alone. You can get there by applying the negation of the definition (3), at which point you've got the desired contradiction. (Do you see why?) However, we can proceed more directly.
Use (2) with the fact that $\epsilon>0$ to conclude that there is some $N_1\in\Bbb N$ such that if $n>N_1,$ then $a_n\in(l_1-\epsilon,l_1+\epsilon).$
Choosing such $\epsilon$ that (1) is valid, from (3) follows: If $\lim a_n = l_2$ then $\lim a_n \ne l_1$.
Again, (3) is not enough to draw this conclusion. So, we instead conclude from (3) that there is some $N_2\in\Bbb N$ such that if $n>N_2,$ then $a_n\in(l_2-\epsilon,l_2+\epsilon).$
Now, what happens if we put $N=\max\{N_1,N_2\}$ and take $n>N$?
From this we conclude that $\lim a_n = l$ is a unique number. This contradicts the assumption that $l_1 \ne l_2$.
You could, indeed, conclude (once you've adequately justified the claims you made) that $\lim_{n\to\infty}a_n$ is a unique number, because we'll already have reached a contradiction. Hence, that last sentence is superfluous.
Best Answer
You are working at $x\ge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$\sin p-\sin q=2\sin\left(\frac{p-q}{2}\right)\cos\left(\frac{p+q}{2}\right).$$
In your case you have $p=\sqrt{x}$ and $q=\sqrt{x-T}$, for $x\ge T.$ So,
$$\frac{\sqrt{x}-\sqrt{x-T}}{2}=k\pi,\quad k\in \Bbb Z$$ or $$\frac{\sqrt{x}+\sqrt{x-T}}{2}=\frac \pi 2+k\pi,\quad k\in \Bbb Z$$
Can you finish?