Wikipedia's proof that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous 
Prove that $f : \mathbb{R} \to \mathbb{R}, x \mapsto x^2$ is not uniformly continuous

Now I actually ask this question, as Wikipedia's proof of this seems wrong to me.

Wikipedia's Proof:
$f$ is uniformly continuous on $\mathbb{R}$ if $\forall \epsilon > 0$, there exists a $\delta > 0$ for all $x_1, x_0 \in \mathbb{R}$ such that
$$|x_1 - x_0| < \delta \implies |f(x_1) -f(x_0)| < \epsilon$$
But $$\begin{aligned} 
f(x+ \delta) - f(x) &= x^2 + 2x\delta +\delta^2 - x^2\\
&= 2x\delta + \delta^2\\
&= \delta(2x + \delta)\\
&> \epsilon \ \ \ \text{for suffiently large $x$} \  \ \ \square
\end{aligned}$$ 

Now what I don't get is that if we set $x_1 = x+ \delta$ and $x_0 = x$, then $|x_1 -x_0| = |x+ \delta - x| = |\delta| \not< \delta$, so everything above should not apply as $|x+ \delta - x| \not < \delta$ in the first place.
Is Wikipedia's proof correct? If so, then why would my argument above be incorrect?
 A: You can replace the $|x - y| < \delta$ with $|x - y | \leq \delta$ in the definition of uniform continuity  and not fundamentally change anything.  However, if it makes you feel comfier, you can fix it up by choosing a $\rho < \delta$ and proceed exactly as shown, but with $f(x+ \rho) - f(x)$.

We can establish the equivalence of the two slightly different definitions as follows:
First, if there exists a $\delta$ such that $|x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$, then there exists a $\rho$ such that $|x - y| \leq \rho \implies |f(x) - f(y)| < \varepsilon$.  For example, $\rho = \delta / 2$.
On the other hand, if there exists a $\delta$ such that $|x - y| \leq \delta \implies |f(x) - f(y)| < \varepsilon$, then of course it's also true that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

One can also change the $< \varepsilon$ condition to $\leq \varepsilon$.  To establish this equivalence, first suppose that if we are given any $\varepsilon > 0$, we can find a $\delta$ such that $|x - y| < \delta \implies |f(x) - f(y)|< \varepsilon$.  It follows immediately that $|x-y| < \delta \implies |f(x) - f(y)| \leq \varepsilon$.
On the other hand, suppose that if we are given any $\varepsilon > 0$, we can find a $\delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| \leq \varepsilon$. By choosing a $\delta$ that satisfies this condition for some $\varepsilon' < \varepsilon$, we thus have a $\delta$ such that $|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon$.

I suspect the reason why the strict less-than condition for both $\varepsilon$ and $\delta$ is prefered in analysis texts is because this emphasizes the more general topological definition of continuity.  Namely, the preimage of an open set is open.
A: You miss an $x$ in $\delta(2+\delta)$. It should be, as the Wikipedia article shows:

But $$
{\displaystyle f(x+\delta )-f(x)=2x\delta +\delta ^{2}=\delta (2\color{blue}{x}+\delta )\ ,}\tag{1}
$$ 
  and for all sufficiently large $x$ this quantity is greater than  $\epsilon$ .


[Added after OP put the $x$ back.]  
The function $f$ being not uniformly continuous is the same as the following:

there exist $\epsilon>0$ such that for any $\delta>0$ there exist $x_1,x_2\in\mathbb{R}$ with $|x_1-x_2|<2\delta$ and 
  $$
|f(x_1)-f(x_2)|\geq \epsilon
$$

A: If we put
$$f(x)=x^2,$$
$$x_n=\sqrt{n+1},$$
and $$y_n=\sqrt{n}.$$
then we have
$$\lim_{n\to+\infty}(x_n-y_n)=0$$
but
$$|f(x_n)-f(y_n)|=1$$
So, $\;\;f\;\;$is not uniformly continuous at $[0,+\infty)$.
If we take $\epsilon=0.5$ for example,
then $\forall \eta>0$,
we will always find $x_n$ and $  y_n$ such that
$|x_n-y_n|<\eta$ and
$|f(x_n)-f(y_n)|=1>\epsilon$.
