[proofplan] We use the $\varepsilon$-definition of [uniform convergence](/page/Uniform%20Convergence) with tolerance $\varepsilon/2$. The resulting index $N$ controls both $d_Y(f_n(x),f(x))$ and $d_Y(f_m(x),f(x))$ uniformly in $x \in E$. The triangle inequality in the [metric space](/page/Metric%20Space) $(Y,d_Y)$ then bounds $d_Y(f_n(x),f_m(x))$ by the sum of these two uniformly small terms. [/proofplan] [step:Choose a uniform convergence index for half the tolerance] Let $\varepsilon>0$ be given. Since $(f_n)_{n=1}^{\infty}$ converges uniformly to $f$ on $E$, applied with the positive number $\varepsilon/2$, there exists $N \in \mathbb{N}$ such that for every $k \ge N$ and every $x \in E$, \begin{align*} d_Y(f_k(x),f(x))<\frac{\varepsilon}{2}. \end{align*} [guided] Let $\varepsilon>0$ be fixed. To prove uniform Cauchyness, we must eventually make $d_Y(f_n(x),f_m(x))$ smaller than $\varepsilon$ using a single index $N$ that works for every $x \in E$ and every pair $m,n \ge N$. The hypothesis gives exactly this kind of uniform index, but for the distance from $f_k(x)$ to the limiting value $f(x)$. We apply uniform convergence with tolerance $\varepsilon/2$. Because $\varepsilon>0$, the number $\varepsilon/2$ is also positive. Hence there exists $N \in \mathbb{N}$ such that for every $k \ge N$ and every $x \in E$, \begin{align*} d_Y(f_k(x),f(x))<\frac{\varepsilon}{2}. \end{align*} This is the only index we will use for both tails of the sequence. [/guided] [/step] [step:Apply the triangle inequality through the uniform limit] Let $m,n \ge N$ and let $x \in E$. Since $f_n(x)$, $f(x)$, and $f_m(x)$ are elements of the metric space $Y$, the triangle inequality for $d_Y$ gives \begin{align*} d_Y(f_n(x),f_m(x)) \le d_Y(f_n(x),f(x)) + d_Y(f(x),f_m(x)). \end{align*} By symmetry of the metric $d_Y$ and by the choice of $N$ applied first to $k=n$ and then to $k=m$, \begin{align*} d_Y(f_n(x),f(x))<\frac{\varepsilon}{2} \end{align*} and \begin{align*} d_Y(f(x),f_m(x))=d_Y(f_m(x),f(x))<\frac{\varepsilon}{2}. \end{align*} Therefore \begin{align*} d_Y(f_n(x),f_m(x))<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \end{align*} [/step] [step:Conclude the uniform Cauchy condition] The index $N$ depends only on $\varepsilon$ and not on $m$, $n$, or $x$. Thus for every $m,n \ge N$ and every $x \in E$, \begin{align*} d_Y(f_n(x),f_m(x))<\varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, $(f_n)_{n=1}^{\infty}$ is uniformly Cauchy on $E$. [/step]