Show that $f_n \to f$ uniformly on $X$ as $n\to \infty$ if and only if for every sequence $x_n$ in $X$ we have $$ dY (f_n(x_n), f(x_n)) \to 0 as n \to \infty.$$
I am wondering if $dY (f_n(x_n), f(x_n))$ means the distance between the point "$f_n(x_n)$" and "$f(x_n)$", or the distance between two sequence?
I think $f_n$ uniformly convergent implies that for every $\varepsilon > 0$, there exists $N$ such that $dY (f_n(x_n), f(x_n)) < \varepsilon$, for all $x \in X$. And since every element $x'$ of any sequence $(x_n)$ is in $X$, $dY (f_n(x_n), f(x_n)) \to 0$ as $n \to \infty$.
But I do not know how to prove conversely. I can prove $f_n$ is pointwisely convergent by constructing $x_n = x, x, x\cdots$ for every $x \in X$, but that cannot prove uniform convergence since $N$ depends on $x$.
Thanks for any help in advance.
Suppose $f_n$ does not converge to $f$ uniformly. Then there exists $\epsilon>0$ such that for each $n\in \mathbb{N}$ there is a $k_n\geq n$ and $y_n\in X$ with $d_Y(f_{k_n}(y_n), f(y_n))\geq \epsilon$. Define a sequence $\{x_n\}_n$ by putting $x_{k_n}=y_n$ for each $n$, and for $k\in \mathbb{N}\backslash\{k_n : n\in \mathbb{N}\}$ just let $x_k$ be whatever. Then $d_Y(f_{k_n}(x_{k_n}), f(x_{k_n}))\geq \epsilon$ for each $n$ so $\{d_Y(f_{n}(x_{n}), f(x_{n}))\}_n$ does not tend to $0$.