My working: $$ x \neq x + kq $$
I solved for kq and got $$ kq \neq 0 $$
Now I'm partially confused because the question had some further detail to it which was :
" Be careful not to assume what you need to prove. Don't start your proof by assuming there's a choice of $q$ where $x = x + kq$. "
After reading that statement I changed my approach. So I decided to negate the theorem and reached to $x≠x+kq$.
I'm not sure if what I did is right, because the last time I did proofs was when I was 16 and now I'm starting to get back into them.
Also if u think it's wrong, please feel free to provide hints. But don't write the answer :) .
Elaborating and filling in the way you wanted to prove it using contradiction.
You start by assuming $x \not\equiv_k x$ and then you wrote $x \neq x + kq$.
However, make sure you actually understand what you mean by that. To be more specific, you have introduced a new character $q$ without any introduction as to what it is.
What you really should write is: $x \neq x + kq$ for any $q \in \mathbb{Z}$.
In that case, the contradiction becomes clear when you rearrange the equation to write: $$kq \neq 0 \quad \text{for } \textbf{any } q \in \mathbb{Z}.$$
Do you now see what the clear contradiction is?
Note: to show that the above is a contradiction, you just need to come up with one value of $q \in \mathbb{Z}$ such that $kq \neq 0$ is not true.
However, I would suggest taking the more direct route of proof by showing that $k \mid (x - x)$.