First, two definitions:
An elementary deformation of knot $P$ is defined as the knot $P'$ if one is determined by the vertices $(p_1, p_2, ..., p_n)$ and the other the vertices $(p_0, p_1, p_2,..., p_n)$ where $p_0$ is not collinear with $p_1$ and $p_n$ and the triangle spanned by $(p_0,p_1,p_n)$ intersects $[p_1,p_2]\cup...\cup[p_{n-1},p_n]\cup[p_n,p_1]$ only in $[p_1,p_n]$.
Knots $P$ and $L$ are equivalent if there is a sequence of knots $P_0, P_1, ..., P_n = L $ such that $P_{i+1}$ is an elementary deformation of $P_i \ $ , $ \ 0 \leq i \leq n-1$.
The problem:
Already shown is that if I move the vertices of knot $P$ a distance less than $\epsilon >0$ to obtain knot $Q$ then $P$ is equivalent to $Q$ in the sense above.
I now need to show that if $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ is given by $f(x)= x + a$ for vector $a$, then $f(L)$ is equivalent to knot $L$.
If you want to use what you've already done, simply measure $a$ to have length $\lVert a\rVert_2 = t$. If $t=0$ we are already done, if not let
be the unit vector in $a$'s direction. Then with $s = [2t\epsilon^{-1}]$ move the knot by ${\epsilon \over 2}\cdot \mathbf{v}$ exactly $2s$ times, then take the last little leap of $(t\epsilon^{-1} - [t\epsilon^{-1}])\cdot \mathbf{v}$. Since the distance each time is less than $\epsilon$ we have found a sequence of knots
with the desired properties.