On bi-invariant metrics on groups I'm looking for a quick, snap-your-fingers proof of the following result:

A continuous length metric on $\mathbb{R}^n$ that is invariant under translations comes from a norm.

To be clear about the terminology: I'll say that a metric $d$ comes from a norm if $d(x,y) = \|x - y \|$. 
I'm also looking for a reference where the following question is addressed:

Given a continuous bi-invariant length metric on a Lie group, is it true that one-parameter subgroups are geodesics and/or that the metric comes from a bi-invariant continuous Finsler metric? 

Remark 1. Please pay attention to the level of generality. These things are really easy  if you assume a bit more regularity than one should.
Edit: May 12, 2014. I had written earlier that I had a proof which used
convolution and the action of the affine or linear group on translationn-invariant distances to regularize the metric. In fact, this approach is wrong or incomplete. It is not clear (or not true?) that the regularized metric is a length metric.
On the other hand, the OP does follow from the much, more general results of Berestovskii cited by Yves in his comment.
 A: Here is a nice quick proof if one assumes that every two points in $(\mathbb{R}^n,d)$ have a unique metric midpoint.

A continuous length metric on $\mathbb{R}^n$ that is invariant under translations and with unique midpoints comes from a norm.

What follows is a simplified presentation of ideas found in Busemann's Geometry of Geodesics (Sections 17 and 50).
Sketch of the proof. The point of departure will be (Busemann's?) characterization of metrics on $\mathbb{R}^n$ that come from norms: 
A complete, continuous metric on $\mathbb{R}^n$ comes from a norm if and only if
the affine midpoint of any two points is also a metric midpoint. 
Recall that a point $b$ is a metric midpoint of two points $a$ and $c$ if
$d(a,b) = d(b,c) = d(a,c)/2$.
It is a standard fact that any two points in a complete length metric space have at least one metric midpoint. Also notice that under the hypothesis of translation invariance completeness comes for free from the fact that $\mathbb{R}^n$ is locally compact. 
If we assume that the metric midpoint is unique, then the proof of theorem
is very simple: Let $a$ and $c$ be any two distinct points in $\mathbb{R}$ and
let $b$ be their metric midpoint. Note that by translation invariance
$$
d(a, a + (c-b)) = d(a + (b-a), a + (c-b) + (b-a)) = d(b,c) = d(a,c)/2 
$$
and
$$
d(a + (c-b), c) = d(a + (c-b), a + (b-a) + (c-b)) = d(a, a + (b-a)) = d(a,b),
$$
which is also $d(a,c)/2$. It follows that the point $a + (c - b)$ is the metric centre of $a$. The upshot: the affine midpoint of two arbitrary points $a$ and $c$, $(a + c)/2$ equals their metric center and the metric must come from a norm.
Remark. The proof shows that the map $x \mapsto a + c - x$ maps the set of metric midpoints of the pair $(a,c)$ into itself. It is also easy to see that this map is an isometric involution. If we could prove that this involution has a fixed point, we would have the general result (i.e., without assuming the uniqueness of metric midpoints).
