[proofplan] The proof uses the defining distance-preservation property of an isometry. Fixing a point $x\in X$, we choose the same radius in the domain as the requested radius in the codomain, namely $\delta=\varepsilon$. The equality of distances then gives the metric-space $\varepsilon$-$\delta$ condition for continuity at $x$. The same equality immediately implies the $1$-Lipschitz inequality. [/proofplan] [step:Use distance preservation to prove continuity at an arbitrary point] Fix $x\in X$ and let $\varepsilon>0$. Define $\delta:=\varepsilon$. Let $y\in X$ satisfy \begin{align*} d_X(x,y)<\delta. \end{align*} Since $f$ is an isometry, it preserves distances between all pairs of points in $X$, so \begin{align*} d_Y(f(x),f(y))=d_X(x,y)<\delta=\varepsilon. \end{align*} Thus for this fixed $x\in X$ and every $\varepsilon>0$, there exists $\delta>0$ such that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\varepsilon$. Therefore $f$ is continuous at $x$. Since $x\in X$ was arbitrary, $f$ is continuous on $X$. [guided] We prove continuity using the metric-space $\varepsilon$-$\delta$ definition. Fix a point $x\in X$. To show that $f$ is continuous at $x$, we must prove that for every $\varepsilon>0$ there exists a number $\delta>0$ such that every $y\in X$ satisfying $d_X(x,y)<\delta$ also satisfies \begin{align*} d_Y(f(x),f(y))<\varepsilon. \end{align*} Let $\varepsilon>0$ be given, and define $\delta:=\varepsilon$. This choice is valid because $\varepsilon>0$, so $\delta>0$. Now take an arbitrary point $y\in X$ such that \begin{align*} d_X(x,y)<\delta. \end{align*} The reason this choice of $\delta$ works is exactly the isometry hypothesis: for every pair of points in $X$, the distance between their images in $Y$ equals the original distance in $X$. Applying that hypothesis to the pair $x,y\in X$, we obtain \begin{align*} d_Y(f(x),f(y))=d_X(x,y). \end{align*} Combining this equality with the assumption $d_X(x,y)<\delta$ and the definition $\delta=\varepsilon$ gives \begin{align*} d_Y(f(x),f(y))=d_X(x,y)<\delta=\varepsilon. \end{align*} This verifies the $\varepsilon$-$\delta$ condition for continuity at the fixed point $x$. Since no special property of $x$ was used beyond $x\in X$, the same argument applies at every point of $X$. Hence $f:X\to Y$ is continuous. [/guided] [/step] [step:Read the Lipschitz constant from the same distance identity] Let $x,y\in X$. Since $f$ is an isometry, \begin{align*} d_Y(f(x),f(y))=d_X(x,y)\leq 1\cdot d_X(x,y). \end{align*} Therefore $f$ is $1$-Lipschitz. Combining this with the preceding step proves the theorem. [/step]