Let $D$ be a compact subset of $\mathbb{R}$ and let $f$ be a shrinking map defined on $D$, show that $f$ has a unique fixed point.
This is a different version of the contracting map theorem in which $D$ is simply closed but $f$ is contracting instead. In this case, I took a sequence $x_{n}=f(x_{n-1})$, showed that it was Cauchy and therefore converges, since $D$ is closed it converges in $D$ and by the continuity of contracting functions, it can be shown that $f(l)=l$ where $l$ is the limit of $x_{n}$.
However, in this case, I seem to be encountering a problem.
Let $x_{n}=f(x_{n-1})$ be a sequence in $D$, this is possible because $f(D)\subseteq D$ for a shrinking map. Now, since $D$ is compact, $(x_{n})_{n}$ has a convergent subsequence in $D$. Let us call it $(x_{n_{k}})_{k}$: $\exists\space l\in D$ such that $\lim_{k\to\ +\infty}x_{n_{k}}=l$
$x_{n_{k}}=f(x_{n_{k}-1})$, so the limit of $f(x_{n_{k}-1})$ is also $l$
By the continuity of $f$ on $D$, the limit of $f(x_{n_{k}})$ is $f(l)$
Now all I need to do is show that $\lim f(x_{n_{k}})=\lim f(x_{n_{k}-1})$ but this is not necessarily true since $n_{k}-1$ may not be one of the $n_{k}$'s, so nothing guarantees that they have the same limit. I'm therefore stuck now. Any tips on what to do?
Edit: definition of shrinking map: Let $D\subseteq\mathbb{R}$ and $f:D\rightarrow D$, then f is shrinking if and only if $\forall\space x\ne y\in D$, $|f(x)-f(y)|<|x-y|$
Existence of a fixed point : Let $g : x \in D \mapsto d(x,f(x))$. Then $g$ is continuous and positive on $D$, which is compact. There exist $a \in D$ such that $g$ is minimal at $a$. Ad absurdum, assume that $f(a) \neq a$. Then $g(f(a)) = d(f(a),f(f(a)) < d(a,f(a)) = g(a)$: absurd. Thus $f(a)=a$.
The unicity is obvious from the definition (take $a,b$ two different fixed points and note that $d(a,b)=d(f(a),f(b)) < d(a,b)$).