Choosing $\epsilon, x, y$ to prove $f(x) = x^2$ is not uniformly continuous on $\mathbb{R}$. In my analysis class we discussed this proof by stating the negation of what it means to be uniformly continuous. The negation is: $\exists \epsilon > 0$ $\forall \delta > 0$ s.t. $\exists x, y \in \mathbb{R}$ where $|x-y| < \delta$ and yet $|f(x) - f(y)| \geq \epsilon.$ So, we have to specify such $\epsilon$, $x$, and $y$. My professor tried to explain the intuition behind choosing $\epsilon = 1$, $x = \frac{1}{\delta}$, and $y = \frac{1}{\delta} + \frac{\delta}{2}$ using the fact that $f'(x) = 2x$, but I still don't really see how this helps. 
I understand how to prove this statement after making these choices, but what's the idea we use to arrive at them?
 A: The idea that we use to arrive at the following values of $\epsilon , x$ and $y$, is the mean value theorem. 
The mean value theorem states that if a function $f$ is continuous on a closed interval $[a,b]$ and differentiable on $(a,b)$, then there is a point $c \in (a,b)$ such that $f(b) - f(a) = f'(c) \times (b-a)$.
Now, suppose that $\min_{c \in (a,b)}|f'(c)| = m$. Taking absolute values on the previous statement, we get $f(b) -f(a) \geq m \times |b-a|$, or that $\frac{|f(b) -f(a)|}{|b-a|}> m$.
You can see that $x^2$ is differentiable on $\mathbb R$, so the MVT can be applied on any interval with this function.

Now, we want points $x,y$ such that $|f(x) - f(y)| > \epsilon$, but $|x-y| < \delta$, for each $\delta$, and some prefixed $\epsilon$. Taking the ratio, we get $\frac{|f(x) - f(y)|}{|x-y|} > \frac \epsilon \delta$. I do not want to fix $\epsilon = 1$ right now, because as you will see, the choice of $\epsilon$ is flexible.
Now, looking at the mean value theorem, candidates for $x$ and $y$ can be decided as follows : if possible, pick $x < y$ from an interval where the minimum value of the derivative is greater than some $m > \frac{\epsilon}{\delta}$, whose value we know, and ensure that $|x-y| < \delta$, while also ensuring that $m |x-y| > \epsilon$. Then, applying the mean value theorem will give us that $\frac{|f(x) - f(y)|}{|x-y|} > m$, and hence $|f(x) - f(y)| > \epsilon$ although $|x-y| < \delta$. (If such a choice is not possible, then we cannot use MVT for the purposes of disproving uniform continuity)

So what works out? Well, the derivative of $f(x)  = x^2$ is $2x$, which is unbounded! Recall what we want again : we want to ensure that $|x-y| < \delta$ and $m = \min_{r \in [x,y]} 2r > \frac{\epsilon}{\delta}$. But $2r$ is increasing, so the minimum is equal to $2x$ which we want greater than $\frac{\epsilon}{\delta}$. So let's just take $x = \frac{\epsilon}{\delta}$ : it is obvious that the derivative condition is satisfied with $m = \frac{2\epsilon}{\delta}$. For the other condition, take $y = \frac{\epsilon}{\delta} + \frac\delta 2$. Now, from MVT we get $\frac{|f(x) - f(y)|}{|x-y|} > \frac{\epsilon}{\delta}$, and we also have $|x-y| = \frac \delta 2$, hence we get $m |x-y| > \epsilon$.

Thus the argument against uniform continuity actually works for any $\epsilon > 0$. Set $\epsilon = 1$  to see what your professor argued.
A: Since you understand how to do the proof but feel it is unintuitive, consider a simpler counterexample.  
The intuition behind nonuniform continuity is that it always possible to find points $x$ and $y$  that can be arbitrarily close but for which the function values $f(x)$ and $f(y)$ cannot be closer than some prescribed tolerance.  If for example, we choose sequences $x_n = n$ and $y_n = n + 1/n$, then as $n \to \infty$ we have $|x_n -y_n| = 1/n \to 0$, but 
$$|x_n^2 - y_n^2| = |x_n - y_n||x_n + y_n| = 2 + 1/n^2 \not\to 0$$
Regarding the role of the derivative (if it exists) and uniform continuity, there is a deeper principle involved here.  
A bounded derivative implies uniform continuity, but there are functions with unbounded derivatives that are not uniformly continuous on unbounded domains.  An example is would be $f(x) = \cos(x^3)/x$ on $[1,\infty)$.
On the other hand, if $\lim_{x \to \infty} f'(x) = +\infty$, as is the case for $f(x) = x^2$ then there can be no uniform continuity on $[1,\infty)$.  Choosing any $x$ and $y = x + \delta/2$, we have $|x-y| = \delta/2 < \delta$. However by the mean value theorem there is a point $\xi \in (x,y)$ such that $|f(x) - f(y)| = |f'(\xi)||x-y|$. For all sufficiently large  $x $ we have $f'(\xi) > 2/\delta$ and 
$$|f(x) - f(y)| > \frac{2}{\delta}\frac{\delta}{2} = 1$$
Thus a derivative tending to $+\infty$ makes it impossible to make $|f(x) - f(y)| < \epsilon < 1$ for every $x,y \in [1,\infty)$ that are closer than any distance $\delta$, no matter how small.
