Preimage of convergent sequence under isometric embedding

Question

Suppose $f: (X,d_X) \to (Y,d_Y)$ is an isometric embedding between two metric spaces. Let $(x_n)$ be a sequence in $X$ such that its image $(y_n)$ converges in $Y$ to a point $y$. Is that enough to infer that there must be an $x\in X$ such that $x_n \to x$ and $f(x) = y$?

Answer 1

No. Consider $X = (0, 1) \subseteq [0, 1] = Y$ with $f = \operatorname{id}_X$ and $x_n = \frac{1}{n}$.