In our lecture we stated that:
For $f: \mathbb R \to \mathbb R, f(x)=x$, i.e. the identity map. The metric space $(\mathbb R, d_{f})$ is then complete, whereby $d_{f}(x,y)=|f(x)-f(y)|$. We wrote this without explanation and I am quite confused as to why this is a given.
I know that, by definition, a metric space is complete iff every cauchy sequence $(x_{n})_{n} \subset \mathbb R$ converges in $\mathbb R$.
First point of confusion is how do I construct any sequence in $\mathbb R$ with respect to an identity map?
Second, am I right in saying that the when I look at the sequence $(x_{n})_{n}$ I am looking at the indices $n$ that lie in the Domain of $f$, while $x_{n}$ is in the range of $f$?
Third, how do I prove that the identity map is indeed complete with regards to the aforementioned metric.
I'm assuming you know that ${\mathbb R}$ with its usual metric $d(x,y):=|x-y|$ is a complete metric space.
Now your "new" metric $d_f$ is just the ordinary "old" metric $d$, since $$d_f(x,y):=\bigl|f(x)-f(y)\bigr|=|x-y|=d(x,y)$$ for all $x$ and $y$. It is then absolutely obvious that ${\mathbb R}$ is also complete with respect to $d_f$.
I suggest that you check your notes: whether the $f$ in question is really the identity map. The setup described in your question seems to be pointless.