Prove that if two miles are run in 7:59 then one mile MUST be run under 4:00.

Consider when the runner passes the one-mile mark. If it is before 4:00 then he ran the first mile in less than 4 minutes. If it is after 3:59, then the second mile was covered in less than 4 minutes. But because 3:59 comes before 4:00 at least one of these cases (and possibly both) must be true.

If $x \geq 4$ and $y \geq 4$ then $x+y \geq 8.$

EDIT: on André's extra credit problem, use Beni's way of writing, time function $f,$ then define $g(m) = f(m+1) - f(m)$ with $0 \leq m \leq 1.$ We know $f(0) = 0, \; f(2) = 8.$ So, $g(0) + g(1) = 8.$ If both $g(0), g(1)$ are $4,$ we are done with André's problem. If one of the pair is above 4, the other is below 4. So, by the Intermediate Value Theorem, there is then some other $0 < m < 1$ such that $g(m) = 4.$

The comment by André Nicolas is related to a very pretty theorem that deserves to be much better known. I first came across it in R.P. Boas's Traveler's Suprises, which appeared in The Two-Year College Mathematics Journal, 10 no. 2 (1979), pp. 82-88 (though I read it in the reprint that appeared in the highly recommended Lion hunting and other mathematical pursuits).

After giving the standard Mean Value Theorem problem ("if you travel in a differentiable way and your average speed is $50$ mph, is there some instant during which your instantaneous speed was exactly $50$ mph?") and a small variant ("Is there some small interval of time during which your average speed was also exactly 50 mph?"), he then gives two very similar but subtler questions:

1. If you travel for time $h$, more than one hour, and average 50 mph for the trip, is there necessarily some one continuous hour during which you covered exactly $50$ miles?

2. Suppose you run a considerable number $m$ of miles and average $8$ minutes per mile. Is there necessarily some one continuous mile (like a "measured mile" on a highway) that you covered in exactly $8$ minutes?

Boas provides the answers by using the Universal Chord Theorem. The theorem was first proven by Levy as a generalization of Rolle's Theorem in Sur une Généralisation du Théorème de Rolle, C. R. Acad. Sci., Paris, 198 (1934) 424-425.

If $f(x)$ is a continuous function, a horizontal chord of $f$ means a line segment with slope $0$ with both ends on the graph of $f$.

Here is Levy's version; you can easily see the similarity with Rolle's Theorem:

Theorem. The values $\alpha = 1, \frac{1}{2},\frac{1}{3},\ldots$ are precisely those for which the following statement is true:

If $f$ is a continuous, real valued function on the closed unit interval such that $f(0)=f(1)$, then the graph of $f$ has a horizontal chord of length $\alpha$.

Proof. Let $f$ be a continuous function defined on $[0,1]$, with $f(0)=f(1)$. Suppose $n$ is an integer greater than $1$, but that $f$ has no horizontal chord of length $\frac{1}{n}$. Let $g(x) = f(x+\frac{1}{n}) - f(x)$; then $g(x)$ is never zero on $[0,1-\frac{1}{n}]$, but is continuous, so it cannot change sign on that interval. If $g(x)\gt 0$, then $$\begin{align*} g\left(1 - \frac{1}{n}\right) &\gt 0\\ g\left( 1 - \frac{2}{n}\right) &\gt 0\\ &\vdots\\ g\left(1 - \frac{n}{n}\right) = g(0)&\gt 0 \end{align*}$$ which is equivalent to $$\begin{align*} f(1) - f\left(1 - \frac{1}{n}\right)&\gt 0\\ f\left(1-\frac{1}{n}\right) - f\left(1-\frac{2}{n}\right) &\gt 0\\ &\vdots\\ f\left(\frac{1}{n}\right) - f(0) &\gt 0. \end{align*}$$ Adding all of them up we conclude that $f(1)-f(0)\gt 0$, a contradiction. If $g(x)\lt 0$, a similar argument implies $f(1)-f(0)\lt 0$.

Therefore, $g(x)=0$ for some $x\in [0,1-\frac{1}{n}]$, but this implies that $f(x+\frac{1}{n}) =f(x)$; that is, the graph of $f$ has a chord of length $\frac{1}{n}$.

(Aside: You'll note this is exactly the argument Will Jaggy gives in his answer.)

Now assume that $\frac{1}{\alpha}$ is not an integer. Then Levy gives the following function: $$f(t) = t\sin^2(\alpha\pi) - \sin^2(\alpha\pi t).$$ Then $f(1)=f(0)$, but an easy calculation shows $f(t +\alpha) - f(t) = \alpha\sin^2(\alpha\pi)\neq 0$ for all $t$. $\Box$

If we assume that the travel in question 1 is continuous, and we assume that the travel in question 2 is such that time is an increasing continuous function of distance (no stopping, no backing up), then one can deduce the affirmative answer to both questions from the Universal Chord Theorem by using the same trick that one uses to deduce the Mean Value Theorem from Rolle's Theorem. I'll leave that derivation to the interested reader.

Question 2 also has an affirmative answer if we drop the assumption that the travel did not include stopping or backing up, but according to Boas the proof is much more difficult (it comes from a generalization of the Universal Chord Theorem proven by Hopf, that any continuous curve that has a chord of length $1$ has a parallel chord of length $\frac{1}{n}$ when $n$ is an integer.; he cites Hopf's Über die Sehnen ebener Kontinuen und die Schleifen geschlossener Wege. Comment. Math. Helv. 9 (1937), 303-319.

Proof by contradiction: If he/she had needed more than (or exactly) 4 minutes for the first and for the second mile, then he/she would have needed more than (or exactly) 8 minutes for both.

You could model your case as follows: denote $f$ the time function, $f:[0,2] \to \Bbb{R}_+$, increasing, with $f(0)=0$, which shows the time at distance $x$. You know that $f(2)=7:59$.

If you suppose that $f(x+1)-f(x)\geq 4:00$ then $f(2)=f(2)-f(1)+f(1)-f(0) \geq 8:00$, and this is a contradiction.