I was asked to check if $f_n(x)=\begin{cases} nx & \text{ if } 0\leq x \leq \frac{1}{n}\\1& \text{ if } \frac{1}{n}<x\leq 1\\ \end{cases}$ uniformly convergence on $\left [ 0;1 \right ]$ by using the definition of uniform convergence or theorem of the uniform distance between bounded functions.
This is my work:
First, I try to find out the pointwise limit function of $f_n(x)$.
For $x=0,\forall n\in\mathbb{N}^*:f_n(0)=n.0=0$, so $\lim_{n\rightarrow\infty}f_n(0)=\lim_{n\rightarrow\infty}0=0.$ For $0<x\leq 1,\forall n\geq\frac{1}{x}:f_n(x)=1$, so $\lim_{n\rightarrow\infty}f_n(x)=\lim_{n\rightarrow\infty}1=1.$
Thus, $f_n(x)$ converges pointwise to the function $f(x)=\begin{cases} 1&\text{if}\: 0<x\leq\ 1\\0 &\text{if}\: x=0 \end{cases}$ on $[0;1]$.
Now I check if $f_n(x)$ uniformly converges to $f(x)$ or not.
We have $$|f_n(x)-f(x)|=\begin{cases} |0-0| & \text{ if } x=0 \\ |nx-1| & \text{ if } 0<x\leq\frac{1}{n} \\ |1-1|& \text{ if } \frac{1}{n}<x\leq 1 \end{cases}=\begin{cases} 0 & \text{ if } x=0\:\text{or}\: \frac{1}{n}<x\leq 1 \\ 1-nx & \text{ if } 0<x\leq\frac{1}{n} \end{cases},$$ so $\underset{x\in[0;1]}{sup}\left | f_n(x)-f(x) \right |=\underset{}{sup}\{\underset{x\in(0;\frac{1}{n}]}{sup}(1-nx);\:0\}=sup\{1;0\}=1.$
(More detail, $\underset{x\in(0;\frac{1}{n}]}{sup}(1-nx)=1$ since $g(x)=1-nx\:$ is monotonically decreasing on $\left(0;\frac{1}{n}\right]$, so $g(x)<g(0)=1-n.0=1, \forall x\in\left(0;\frac{1}{n}\right] $)
We see that $\lim_{n\rightarrow\infty}\underset{x\in[0;1]}{sup}\left | f_n(x)-f(x) \right |=lim_{n\rightarrow\infty}\: 1=1\ne 0$. Thus, $f_n(x)$ does not uniformly converges to $f(x)$ on $[0;1]$.
Is my work correct? What I need to fix and modify? The way I find $\underset{x\in[0;1]}{sup}\left | f_n(x)-f(x) \right |$ is correct? This is I'm not quite sure!
And another question, if I'm not required to do this by using the definition of uniform convergence or theorem of the uniform distance between bounded functions. Can I say something like this : Since $f_n(x)$ is continuous on $[0;1]$ while $f(x)$ is not, so $f_n(x)$ does not uniformly converges to $f(x)$ on $[0;1]$?
I don't see any problem on your proof. As you say at the end, there is a simpler way to see that the convergence is not uniform using the following result: If a sequence $(f_n)$ of continuous functions converges uniformly on $A\subset\mathbb{R}$, then the limit is a continuous function on $A$. Since the functions $f_n$ are continuous for every $n$, and the pointwise limit is discontinuous, then the convergence cannot be uniform.