If i show that $x+a=x+b$ only if $a=b$, does that prove that the above is also true?
$ x+a=x+b \iff x+a-x-b=0 \iff a-b=0 \implies b=a$ also is this any good?
That argument isn't going to work, since at no point are you using any properties of the floor function.
What will work is something like this (only a hint):
If $a \ne b$ then we can assume (without loss of generality) that $a < b$, and then we can take some $y$ such that $ a < y < b$. Can you then find some expression for $x$ which will make sure that $x+a$ and $x+b$ will have different integer parts?
Let [x] be the whole part of x,
[x+a] = [x+b] for every real x => b-1 < [x+a]-x <= b and a-1 < [x+b]-x <= a for every real x this means that b-1 <= a-1 and a-1 <= b-1 so a = b.