I have to prove the following:
Let $(X, d_X)$ and $(Y,d_Y)$ be metric spaces which are proper (i.e. closed balls are compact). If $f \colon X \to Y$ is an isometric embedding such that $f(X)$ is closed in $Y$, then for each $y\in Y$ there is an $x \in X$ such that $d_Y(y,f(x)) = \min_{x\in X} d_Y (y,f(x))$.
Proof: Let $y\in Y$ be arbitrary and choose an $x' \in X$. Next, consider the closed ball $B$ with center $y$ and radius $r=d_Y(y, f(x'))$. Since $B$ is compact and $f(X)$ is closed, $B \cap f(X)$ is compact. Therefore, the continuous function \begin{align*} d_Y(y, f(\cdot)) \colon B \cap f(X) &\to \mathbb{R} \\ f(x) &\mapsto d_Y(y, f(x)) \end{align*} attains its minimum, i.e. there is an $x_\min \in f^{-1} (B)$ such that \begin{equation*} d_Y(y,f(x_\min)) = \min_{x\in f^{-1} (B)} d_Y (y,f(x)). \end{equation*} Also, for any $x \not\in f^{-1} (B)$ we have $d_Y(y,f(x)) > r \geq d_Y(y,f(x_\min))$ and therefore \begin{equation*} d_Y(y,f(x_\min)) = \min_{x\in X} d_Y (y,f(x)). \end{equation*}
Is the proof correct? Could it be shortened?
EDIT: Following jJjjJ's comment, I wrote $B$ instead of $\overline{B}$ so that the closed ball is not confused with the closure of the ball.