[proofplan] We unwind the definitions of uniform continuity and the Cauchy property. Given $\varepsilon > 0$, uniform continuity produces $\delta > 0$ such that close inputs map to close outputs. The Cauchy hypothesis gives $N$ with $d_X(x_m, x_n) < \delta$ for $m, n \ge N$. Composing these two quantitative controls yields $d_Y(f(x_m), f(x_n)) < \varepsilon$ for all $m, n \ge N$. [/proofplan] [step:Fix $\varepsilon > 0$ and extract the modulus of uniform continuity] Let $\varepsilon > 0$ be given. Since $f: X \to Y$ is uniformly continuous, there exists $\delta > 0$ (depending on $\varepsilon$ but not on the choice of points in $X$) such that \begin{align*} d_X(x, y) < \delta \implies d_Y(f(x), f(y)) < \varepsilon \quad \text{for all } x, y \in X. \end{align*} [guided] Let $\varepsilon > 0$ be given. We need to find $N \in \mathbb{N}$ such that $d_Y(f(x_m), f(x_n)) < \varepsilon$ for all $m, n \ge N$. The tool at our disposal is uniform continuity: the defining property of a uniformly continuous function $f: X \to Y$ guarantees that for every $\varepsilon > 0$, there exists $\delta > 0$ such that $d_Y(f(x), f(y)) < \varepsilon$ whenever $d_X(x, y) < \delta$, for all $x, y \in X$. The uniformity is essential here — $\delta$ depends only on $\varepsilon$, not on the particular pair $(x, y)$. This is precisely what distinguishes uniform continuity from pointwise continuity and what makes the argument work: we need a single $\delta$ that works for all pairs $(x_m, x_n)$ simultaneously. Apply the definition of uniform continuity with our given $\varepsilon$ to obtain $\delta > 0$ as above. [/guided] [/step] [step:Use the Cauchy property to make $d_X(x_m, x_n) < \delta$ for large indices] Since $(x_n)_{n=1}^\infty$ is a Cauchy sequence in $(X, d_X)$, applying the Cauchy condition with the value $\delta > 0$ obtained in the previous step produces $N \in \mathbb{N}$ such that \begin{align*} d_X(x_m, x_n) < \delta \quad \text{for all } m, n \ge N. \end{align*} [guided] We now invoke the Cauchy hypothesis. By definition, $(x_n)_{n=1}^\infty$ is Cauchy in $(X, d_X)$ if for every $\eta > 0$ there exists $N \in \mathbb{N}$ with $d_X(x_m, x_n) < \eta$ for all $m, n \ge N$. We apply this definition with $\eta := \delta$ (the value from the previous step) to obtain $N \in \mathbb{N}$ such that $d_X(x_m, x_n) < \delta$ for all $m, n \ge N$. Note that this is the step where merely pointwise continuity of $f$ would fail: a pointwise-continuous function produces $\delta$ depending on the base point, so we would need a different $\delta$ for each pair $(x_m, x_n)$, and there is no guarantee that a single $N$ works for all such pairs simultaneously. [/guided] [/step] [step:Combine the two estimates to conclude $(f(x_n))_{n=1}^\infty$ is Cauchy] For all $m, n \ge N$, we have $d_X(x_m, x_n) < \delta$. By the uniform continuity estimate from the first step, this implies \begin{align*} d_Y(f(x_m), f(x_n)) < \varepsilon \quad \text{for all } m, n \ge N. \end{align*} Since $\varepsilon > 0$ was arbitrary, this shows that $(f(x_n))_{n=1}^\infty$ is a Cauchy sequence in $(Y, d_Y)$. [guided] We chain the two quantitative controls. For $m, n \ge N$: \begin{align*} d_X(x_m, x_n) < \delta \quad &\text{(Cauchy property of $(x_n)$ with threshold $\delta$)} \\ \implies d_Y(f(x_m), f(x_n)) < \varepsilon \quad &\text{(uniform continuity of $f$ with modulus $\delta \mapsto \varepsilon$)}. \end{align*} Since $\varepsilon > 0$ was arbitrary, we have shown: for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that $d_Y(f(x_m), f(x_n)) < \varepsilon$ for all $m, n \ge N$. This is exactly the definition of $(f(x_n))_{n=1}^\infty$ being a Cauchy sequence in $(Y, d_Y)$. It is worth noting why the converse fails — that is, why mere (pointwise) continuity does not preserve Cauchy sequences. Consider $X = (0, 1)$ with the standard metric, $Y = \mathbb{R}$, and $f: X \to Y$ defined by $f(x) = 1/x$. The sequence $x_n = 1/n$ is Cauchy in $X$ (it converges to $0$ in $\mathbb{R}$, hence is Cauchy), yet $f(x_n) = n$ is unbounded and therefore not Cauchy. The function $f$ is continuous but not uniformly continuous on $(0,1)$, precisely because the modulus $\delta$ depends on the base point and degenerates as $x \to 0$. [/guided] [/step]