Lets take the infinite sequence of functions $f_{n}(x) = x/(x +1/n) , x \in [0, 1], n \in \Bbb N$. Show that each function $f_{n}$ is uniformly continuous.
My solution:
Given $\epsilon >0$, let $x, y$ be such that $|x-y|<\delta$. Then, $$ |f_{n}(x)-f_{n}(y)|=\bigg| \frac{x}{x+(1/n)}-\frac{y}{y+(1/n)}\bigg|=\bigg |\frac{x-y}{n[x+(1/n)][y+(1/n)]}\bigg| \leq (x-y)n$$
Choose $n<\epsilon/\delta$, then
$|f_{n}(x)-f_{n}(y)|<\epsilon$. Hence proved.
Is this proof correct, please rectify if any flaw is there. Else suggest a different proof.
Your proof is fine except the $(x-y)$ should be $|x-y|$.
Here is a shorter version: You have $f_n'(x) = {n \over (nx+1)^2}$, and $|f_n'(x)| \le n$, hence the mean value theorem shows that $|f_n(x)-f_n(y)| \le n |x-y|$.