Unconventional (but instructive) proofs of basic theorems of calculus Inspired by this questions asked on MathOverflow, I would like to ask if you know some "sophisticated" proofs of the basic theorems in a calculus course (that is, the ones that you can find, for instance, in Spivak's Calculus). 
In this case, by "sophisticated", I do not mean awfully complicated, but unexpected, extremely clever and unconventional (and hopefully instructive), either because they use concepts from other areas of mathematics or because they enlighten a theorem by tackling it from a non-obvious and by no means standard(ized) perspective.
 A: I think the classic example of this is the whole field of non-standard analysis.  It took 300 years to make infinitesimals rigorous (finally realized in the 1960's), but once equipped with such a toolkit you can derive all the basic calculus results (and much more) in just a couple of lines of infinitesimal algebra. 
A: I always love to prove that:

If $\{a_n\}_{n\in\mathbb{N}}$ is a bounded real sequence, it has a converging subsequence.

with the Erdos-Szekeres', or Dilworth's, theorem:

(Erdos-Szekeres, finite version) Every sequence with $n^2+1$ terms admits a weakly monotonic subsequence with $n+1$ terms.
(Dilworth, infinite version) Every infinite POset contains an infinite chain or antichain.

To prove Erdos-Szekeres, we send $n^2+1$ people to a post office with $n$ employees, $n$ queues.
When a person arrives, he takes place in the first queue such that he is taller than the last person in the queue. If at some point someone ($A$) is not able to take place, then the people in the last position of every queue and $A$ give a decreasing sequence. On the other hand, if everyone is able to take place, there is a queue with at least $n+1$ people in it, giving an increasing sequence.
So we can use Erdos-Szekeres' or Dilworth's theorem to extract a (weakly) monotonic and bounded subsequence from $\{a_n\}_{n\in\mathbb{N}}$. Such a subsequence is clearly converging to its $\sup$ or $\inf$, and we are done.
A: It's interesting to try to prove the basic results on derivatives without using the mean value theorem.
See for instance this previous MO discussion of the role of MVT in first-year calculus, a related post I made on the Expii "Mean Value Theorem" page, and (also on Expii) a MVT-free proof (inspired by the proof (given e.g. in Stein & Shakarchi) of Cauchy/Goursat from complex analysis, following a blog post of Gowers) of the fact that $f'=0$ on an interval implies $f$ is constant.
A: Here is an unconventional but instructive proof of the extreme value theorem for a continuous function $f$ on the interval $[0,1]$. Let $H$ be (gasp) an infinite hypernatural. Partition the interval into $H$ equal subintervals, each of infinitesimal length.  Among the partition points $p_i$, choose the one, say $p_{i_0}$, with the maximal value of $f$. Now round off $p_{i_0}$ to the nearest real number $c$, so that $p_{i_0}$ is infinitely close to $c\in\mathbb{R}$.  Then $c$ is a required maximum of $f$. This is because, by definition of continuity, the composed function $\text{st}\circ f$ is constant on the halo of $c$, where "st" is the standard part function.
