I am trying to prove the following property, which seems fairly intuitive, at least in $\mathbb{R}^n$.
Let $(X,d)$ be a compact metric space where $Y \subseteq X$ arbitrary. Prove that if there exists an isometry $f: \, X \rightarrow Y$, then $X=Y$.
My idea: Prove that $X \subseteq Y$.
My issue at the moment is that I am not completely sure if the existence of an isometry would help us deduce that $X \subseteq Y$.
Another idea that I had was to perhaps use proof by contradiction. Isometries are 'distance preserving' and as $Y$ is a subset of $X$, then there must be two points in $X$ that are 'far enough' so that their images in $Y$ cannot 'match' the distance. However, this only really makes sense when $X$ and $Y$ are finite, which may not necessarily be the case.
If $f[X]$ is proper, pick $x_0 \in X, x_0 \notin f[X]$. Then $d(x_0, f[X]) = r > 0$, as $f[X]$ is also compact, thus closed.
Then define $x_{n+1}= f(x_n)$ for $n\ge 0$, then (as all other $x_n$ besides $x_0$ are $f$-images) we have $d(x_0, x_n) \ge d(x_0, f[X]) \ge r$ for all $n \ge 0$.
Hence, as $f$ is an isometry, $d(x_k, x_{n+k}) = d(x_0, x_n) \ge r$ for all $n \ge 0, k \ge 0$, which can be seen by induction on $k$, fixing $n$.
It follows that for all $n \neq m$, $d(x_n, x_m) \ge r$ which means that $(x_n)$ can have no convergent subsequence, contradicting compactness. So $f[X] = X$.