Can all unit-distance graphs have their vertices at algebraic integers? A graph $G$ is described as a unit-distance graph if there exists a function $f:G \rightarrow \mathbb{C}$ such that for every edge $(u,v) \in E(G)$, we have $|f(u) - f(v)| = 1$.
Obviously, we can necessarily find an embedding into the algebraic numbers $\bar{\mathbb{Q}}$. In other words, if $G$ is a unit-distance graph, we can find $g:G \rightarrow \bar{\mathbb{Q}}$ such that for all $(u,v) \in E(G)$, we have $|g(u) - g(v)| = 1$.
But what about if we further restrict ourselves to the algebraic integers $\mathcal O(\bar{\mathbb{Q}})$? Can all unit-distance graphs be embedded in such a way?
 A: Not an answer to your (interesting!) question.
But this quote from 

Peter Brass, William O. J. Moser, János Pach. Research Problems in Discrete Geometry.
  Vol. 18. New York: Springer, 2005, p.238:

seems tangentially relevant:

 
 
 


in particular, 

all unit-distance graphs in rational $2$- and $3$-dimensional spaces are bipartite.

So at least restricting to rationals limits the realizable unit-distance graphs.
A: $\let\eps\varepsilon$No. I will present a graph whose realization necessarily contains a pair of vertices at distance $1/2$. THis cannot happen if the vertices are algebraic integers.
Firstly, we note that we may force a graph to contain a given piece of triangular lattice. It will be clear after we understand how to enforce the two opposite vertices of a $60^\circ-120^\circ$ unit-sided rhombus to be distinct. This is made by augmenting this rhombus to a Mosers' spindle: the vertices of its realization cannot coincide.
Thus we may enforce two vertices $A$ and $B$ to have any distance realized in the triangular lattice.
Now consider five points $A$, $B$, $C$, $D$, $E$ such that $AB=AC=AD=2$, $BC=DC=BE=DE=1$. If $B\neq D$ and $C\neq E$ then we have $CE=1/2$ as required. Thus it remains to enforce these relations.
To enforce $B\neq D$ it suffices to introduce a point $X$ with $BX=2$ and $DX=1$. To enforce $C\neq E$ it suffices to introduce $Y$ with $CY=2$ and $EY=\sqrt 3$.
REMARK (expanded). Maehara showed that any algebraic number can be realized as a distance between two vertices of a rigid unit-distance framework. This inspired the answer; I just needed to make the graph "more rigid". A similar reinforcing may be applied to any rigid framework to make it "absolutely rigid".
To perform this, for every two points $A$, $B$ at distance $d$ in the fixed rigid realization, one needs to ensure that $d-\eps<AB<d+\eps$ in each realization. 
To ensure $AB>d-\eps$, find in the triangular lattice two distances $\ell_1$ and $\ell_2$ with $d-\eps <\ell_1-\ell_2<d$ and introduce a point with $AX=\ell_1$, $BX=\ell_2$.
Now, similarly to the construction above, one may realize the distance $1/n$ for every $n\in\mathbb N$. Finally, connecting $A$ and $B$ by an appropriate chain of segments of length $1/n$ one ensures that $AB<d+\eps$.
