Show that every graph can be embedded in $\mathbb{R}^3$ Show that every graph can be embedded in  $\mathbb{R}^3$ with all
  edges straight. 


(Hint: Embed the vertices inductively, where should you
 not put the new vertex?)

I've also received a tip about using the curve ${(t, t^2 , t^3 : t \in \mathbb{R} )}$ but I'm not sure how to do that and how I should go about proving this rigorously but without using any pure topology. 

Any hints or useful suggestions?

Thanks a lot!
Let $\vec{p} : \mathbb{R} \to \mathbb{R}^3$ be the function
$\vec{p}(t) = (t, t^2, t^3)$ and $\mathscr{C}$ be the curve 
$\Big\{\;\vec{p}(t) : t \in \mathbb{R}\;\Big\}$.

For any four distinct $t_1, t_2, t_3, t_4$ in $\mathbb{R}$, the volume of the tetrahedron $\mathscr{T}$ formed by
$\vec{p}(t_i) \in \mathscr{C}$ is proportional to a Vandermonde determinant:

$$\begin{align} 
6 \text{Volume}(\mathscr{T}) 
&amp; = \Big| 
(\vec{p}(t_2)-\vec{p}(t_1)) \cdot  \left[
(\vec{p}(t_3) - \vec{p}(t_1)) \times (\vec{p}(t_4) - \vec{p}(t_1)) \right]
\Big|\\
&amp; = \det \begin{bmatrix}
1 &amp; t_1 &amp; t_1^2 &amp; t_1^3\\
1 &amp; t_2 &amp; t_2^2 &amp; t_2^3\\
1 &amp; t_3 &amp; t_3^2 &amp; t_3^3\\
1 &amp; t_4 &amp; t_4^2 &amp; t_4^3
\end{bmatrix} \ne 0
\end{align}$$
The implies any four distinct points on $\mathscr{C}$ are not coplanar. As a result,
the edges of the tetrahedron $\mathscr{T}$ intersect at and only at the appropriate
vertices.

Now take arbitrary $n$ distinct points $\vec{p}_i$ from $\mathscr{C}$. 
Above argument shows that if we form a complete graph $K_n$ from them,
the edges intersect at and only at appropriate vertices. This gives us
an embedding of the complete graph $K_n$ into $\mathbb{R}^3$. 
Since every graph of $n$ vertices is a sub-graph of $K_n$, we are done.
A graph can be seen as a manifold; there is a result that the graph of a smooth function is a manifold; in this case, it is a 1-manifold. Then use Whitney's theorem that guarantees  that a smooth m-manifold can be smoothly-embedded in $\mathbb R^{2m}$.