Suppose $(X, d_X)$ is a complete metric space and $f\colon X \to Y$ an isometric embedding into the metric space $(Y, d_Y)$.
Claim: $f(X)$ is closed in $Y$.
Proof: Let $(x_n)_{n\in \mathbb{N}}$ be a sequence of points in $X$ whose image $(y_n)_{n\in \mathbb{N}}$ converges to a $y \in Y$. Then, $(y_n)_{n\in \mathbb{N}}$ is a Cauchy sequence, and since the embedding is isometric, so is $(x_n)_{n\in \mathbb{N}}$, which therefore converges in the complete space $X$ to an $x \in X$. It follows that \begin{equation*} d_Y(y_n, f(x)) = d_X (x_n, x) \to 0 \end{equation*} and therefore $y = f(x) \in f(X)$.
All correct?