Intersection of nested closed bounded convex sets in Euclidean space

I read that in a complete Euclidean space - i.e. a normed real space with the norm induced by the scalar product - any sequence of nested bounded non-empty closed convex sets has a non-empty intersection, but I can't manage to prove it to myself.

Has anybody any ideas or links to online proofs?

Context

The space is not assumed finite dimensional. At the point where I am in the text, Kolmogorov and Fomin's, they haven't defined a Hilbert space yet.

I have not yet learned about the weak topology and reflexivity.

Let $E_n$, $n\ge 1$, be these nonempty closed convex sets. We want to pick a point $x_n$ in each set so that they form a Cauchy sequence. If the diameter of $E_n$ tends to zero as $n\to\infty$, the Cauchy property comes automatically. Otherwise, we have to make some intelligent choices of $x_n$. I present two versions; the second, pointed out by Daniel Fischer, is more slick.

Close to the "center of the set"

One idea is to pick $x_n$ close to the "center" of each set. To make this precise, let $$r_n = \inf\{r>0: \exists x\in E_n \text{ such that } E_n\subset B(x,r)\}\tag1$$ Here $B(x,r)$ is closed ball of radius $r$ centered at $x$. (The number $r_n$ is sometimes called the Chebyshev radius of $E_n$.) Observe that $r_n$ is a decreasing sequence of positive numbers, so it has a limit, $r_n\to r_*$.

As usual in infinite dimensions, we don't know if the infimum (1) is actually attained. So we must provide some slack: pick $x_n\in E_n$ such that $E_n\subset B(x_n, r_n+2^{-n})$.

I claim that the sequence $(x_n)$ is Cauchy. Indeed, suppose there is $\epsilon>0$ such that there are arbitrarily large indices $n<m$ for which $\|x_n-x_{m}\|\ge \epsilon$. Then $$E_m \subset B(x_n,r_n+2^{-n})\cap B(x_m,r_m+2^{-m})$$ Observe that both radii here can be made arbitrarily close to $r_*$ by choosing $n,m$ large. Using the parallelogram law, you can show that $$B(x_n,r_n+2^{-n})\cap B(x_m,r_m+2^{-m})\subset B\left(\frac12 (x_n+x_m),\rho\right)$$ for $\rho<r_*$, thus arriving at a contradiction.

(Draw a picture of two intersecting balls of nearly the same radius: if the distance between their radii is substantial, the intersection is contained in a ball of smaller radius).

Close to the origin of the space

Another idea is to pick $x_n$ of small norm. Let $$R_n = \inf\{\|x\| : x\in E_n \}\tag2$$ Observe that $R_n$ is an increasing sequence of positive numbers, so it has a limit, $R_n\to R_*$.

Pick $x_n\in E_n$ such that $\|x_n\| < R_n+2^{-n}$.

I claim that the sequence $(x_n)$ is Cauchy. Indeed, suppose there is $\epsilon>0$ such that there are arbitrarily large indices $n<m$ for which $\|x_n-x_{m}\|\ge \epsilon$. Then $$\frac{x_n+x_m}{2}\in E_m$$ by convexity. Using the parallelogram law, you can show that $$\left\|\frac{x_n+x_m}{2}\right\| <R$$ when $m,n$ are sufficiently large, thus arriving at a contradiction.