Here is the problem.
Let $M$ be the metric space of all real numbers, and let $x_0 \in M$. Prove that there exist exactly two isometries of $M$ that leave $x_0$ fixed.
I am having trouble comprehending what it means by that "leave $x_0$ fixed."
Is it asking to prove that there are exactly two isometries $f_1:M \rightarrow M$ and $f_2: M \rightarrow M$ where $f_1(x_0) = x_0$ and $f_2(x_0) = x_0$?
But we don't have a specific metric defined on the set, so any hint on how would I go about proving this?
The intended metric is certainly the usual metric on the real numbers: $d(x,y) = |x-y|$. You are looking to find a function $f$ satisfying $|f(x) - f(x_0)| = |x - x_0|$ which reduces in this case to $|f(x) - x_0| = |x - x_0|$. You can solve this for $f(x)$.