Easy proof of the fact that isotropic spaces are Euclidean
It is a famous problem (due to Banach and Mazur) whether a separable infinite dimensional Banach space which has a transitive isometry group must be isometrically isomorphic to a Hilbert space. Of course, if every two dimensional subspace has a transitive isometry group, then the space is a Hilbert space since then the norm satisfies the parallelogram identity. For counterexamples in the non separable setting, consider the $\ell_p$ sum of uncountably many copies of $L_p(0,1)$ with $p$ not $2$.
For a recent paper related to the Banach-Mazur rotation problem, which contains some other references related to the problem, see
http://arxiv.org/abs/math/0110202.
The heart of the matter is to define a canonical inner product for any norm in finite dimensions. Since it is canonical, an $X$-isometry is also an isometry of the inner product. If the group is transitive on lines, you thus immediately get that norm is Euclidean.
There are two popular ways to do this. One is John's theorem: The ellipsoid in the unit ball $K$ of $X$ (which is any convex, centrally symmetric body) with the largest volume is unique. Or of course you could use John's theorem dually, taking the smallest ellipsoid that contains $K$. The other popular, canonical ellipsoid is the Legendre ellipsoid of $K$, by definition the ellipsoid $L$ that has the same moment of inertia matrix as $K$, assuming that both $L$ and $K$ have uniformly distributed mass.
On the other hand, the averaging argument is also "slick" in my opinion, and I don't really see anything wrong with it even for undergraduates. Arguably the problem with any slick argument is that it is too adroit for some students.
There is a classic paper by Jordan and von Neumann where they prove results that allows this question is settled in an elementary way.
On Inner Products in Linear, Metric Spaces Author(s): P. Jordan and J. V. Neumann, The Annals of Mathematics, Second Series, Vol. 36, No. 3 (Jul., 1935), pp. 719-723.
They first prove by elementary arguments (their Theorem I) the so-called Jordan v. Neumann criterion, that a Banach space is Hilbert iff for all $x$ and $y$, $(*) ||x + y||^2 - ||x - y||^2 = 2||x||^2 + 2 ||y||^2$. They then show from this that a Banach space is Hilbert iff every 1 and 2 dimensional subspace is Euclidean. Here is their argument:
4.The condition that every $<= 2$-dimensional subspace $L'$ of $L$ be isometric to a Euclidean space, is obviously necessary for the existence of an inner-product in the generalized linear,metric space L. It is sufficient,too, because if it is fulfilled, one can argue as follows:If $f_o,g_0\in L$ the space $L'$ of all $\alpha f_0 + \beta g_0$ ($\alpha,\beta$ arbitrary complex numbers) is $<= 2$ dimensional,thus (*) holds in $L'$ (as in every Euclidean space). Therefore it holds in particular for $f = f_0, g = g_0$,and as $f_0,g_0$ are arbitrary, Theorem I proves the existence of an inner product.
[SEE BELOW: The following sentence does NOT complete the proof !]
And as rpotrie has pointed out in another answer, the two dimensional case follows from the assumed transitivity condition.
ERROR NOTICE: I noticed a serious error in the above reasoning! If the isometry group $G$ of a Banach space $V$ is transitive on the unit sphere of $V$, it does NOT follow in any obvious way that the isometry group of a subspace $V'$ of $V$ is transitive on the unit sphere of $V'$. (If $e_1,e_2$ are unit vectors in $V'$, then an element $g$ of $G$ that carries $e_1$ to $e_2$ need not leave $V'$ invariant.)
I did not at first realize how remarkable the conclusion is that transitivity on the unit sphere implies Euclidean. It can be rephrased as saying that transitivity on $S$ implies $2$-transitivity, which to me at least seems even more remarkable. (It was realizing this fact that let me see my silly error.)