Hi take a look of my proof and fix if there is wrong or unnecessary part
$F(x) = x^2 , \alpha \in [0,\infty] $ and prove $F(x)$ is not uniformly continuous
anyway
proof
assume that this$ f(x)$ is uniformly continuous. then there should be a some limit point(??) exist $L$, and $$y = L + \epsilon/2 , x = L-\epsilon/2$$ we can write this limit point as $$|x-y| < \epsilon$$
since $f(x)$ is uniformly continuous there should be exit of $$|f(x) - f(y)| < \delta $$
If we proof there exist some $|f(x)- f(y)| > \delta $ this proof is done
$$|f(x) - f(y)| < \delta = |x^2 - y^2| < \delta = |x+y||x-y|$$
we know $|x-y| < \epsilon/2 $ then write $x<y+\epsilon/2 $
$$|x+y| = |y+\epsilon/2 + y |,$$ then $$x = 2(y+\epsilon)$$
Therefore, $$|x+y||x-Y|> \epsilon$$
there is contradiction.
I'm having trouble understanding your argument - what are you getting at with the 'limit point'? You also seem to have the 'usual' role of $\delta$ and $\epsilon$ switched. Of course, this is just notational (and not mathematical), but sticking with conventions will make your work easier to read.
Basically, I'd say you have some idea what needs to be proved and how to prove it, but the biggest problem is you make no indication of what conditions $x$ and $y$ must satisfy. In particular, given an $\delta$ and $\epsilon$, there are going to be some points where $|f(x) - f(y)| < \epsilon$, you have to find a pair where this fails.
To show $f$ is not uniformly continuous, here's an outline.
1) If $f$ is uniformly continuous, then for every $\epsilon > 0$, there is $\delta > 0$ so that, for any $x,y$ such that $|x-y| < \delta$ implies $|f(x)-f(y)| < \epsilon$.
So to show it is not uniformly continuous, we need to show that for some $\epsilon > 0$, for every $\delta > 0$, there exists $x,y$ such that $|x-y| < \delta$, but $|f(x) - f(y)| > \epsilon$.
2) So, start with an $\epsilon$ that is convenient for you (since it has to hold for any $\epsilon$). Try $\epsilon = 1$ to begin with (we can always start over and make it smaller if this doesn't work).
3) Now, choose any $\delta > 0$. We want to find $x,y$ so that $|x-y| < \delta$, but $|x^2 - y^2| > 1$. Intuitively, we want to pick really big (but close together) $x$ and $y$ (close together so that $|x-y|$ is small, but the difference of their squares is big).
Let's set $y = x + \delta/2$ (so automatically, we have $|x-y| < \delta$) and then decide how big $x$ has to be. As you note, $$|x^2 - y^2| = |x + y||x-y| = |x + x + \delta/2||x - (x + \delta/2)| \\ = \frac{\delta}{2}\left|2x + \frac{\delta}{2}\right|\\ > \frac{\delta}{2}\left|2x\right| = \delta|x|$$
So, if we choose $x$ so that $|x| > \dfrac{1}{\delta}$, then we will have $|x^2 - y^2| > 1$ as needed.