All isometric embeddings that maps $\mathbb{R}$ to an inner product vector space $\mathbb{E}$ are of the form Let $\mathbb{E}$ be a normed vector spaces with inner product. 
All isometric embeddings $f:\mathbb{R} \to \mathbb{E}$ are of the form 
$f(t)=u+tv$, with $u,v \in \mathbb{E}$ and $\|v\|=1$. 
This was an assignment due yesterday. 
I kind of used a similar solved problem from the textbook I am using [Espaços Métricos, E. L. Lima], but found out it proved nothing, specially not that ALL isometric embedddings like $f$ are of the same form.  
While trying to find useful information about this I found this paper on arXiv https://arxiv.org/pdf/1202.0503.pdf
In page 6, Lemma A.1 states that 
"In an inner product space $(V, \|k\|)$ a sphere and a straight line can 
coincide in at most two points."

Let $f$ be such an isometry, $u=f(0)$, $v=f(1)-u$. As $\|f(x)-f(y)\|=|x-y|$ for all $x,y\in\Bbb R$, this implies $\|v\|=1$. 
Let $t\in\Bbb R$ be arbitrary.
Let $w=f(t)-(u+tv)$. Then 
$$\begin{align}t^2&=\|f(t)-f(0)\|^2\\&= \langle tv+w,tv+w\rangle\\&=t^2\|v\|^2+2t\langle v,w\rangle+\|w\|^2\\&=t^2+2t\langle v,w\rangle+\|w\|^2\end{align}$$
so that
$$\tag1 2t\langle v,w\rangle+\|w\|^2=0.$$
Also,
$$\begin{align}(t-1)^2&=\|f(t)-f(1)\|^2\\&= \langle (t-1)v+w,(t-1)v+w\rangle\\&=(t-1)^2\|v\|^2+2(t-1)\langle v,w\rangle+\|w\|^2\\&=(t-1)^2+2(t-1)\langle v,w\rangle+\|w\|^2\end{align}$$
so that
$$\tag2 2(t-1)\langle v,w\rangle+\|w\|^2=0 $$
Subtracting $(1)$ and $(2)$, we find $\langle v,w\rangle = 0$.
Plugging that back into $(1)$, we find $w=0$, i.e.,
$$ f(t)=u+tv.$$
