Show that $\underset {x \in [0,1]} {\sup} f_n(x) \rightarrow \underset {x \in [0,1]} {\sup} f(x)$ as $n \rightarrow \infty$. 
Let $\{f_n\}$ be a sequence of continuous functions converging uniformly to a function $f$ on $[0,1]$. Then show that $\sup\limits{x \in [0,1]} f_n(x) \rightarrow \sup\limits_{x\in[0,1]} f(x)$ as $n \rightarrow \infty$.

My attempt $:$
First of all let us state the following result $:$

Let $S$ and $T$ be two bounded subset of $\mathbb R$. Let $A = \{|x-y| : x\in S , y \in T \}$.Then $A$ is also bounded and $\sup A = \sup S - \inf T$.

Now with the help of above theorem we have 
$M_n= \sup\limits_{x \in [0,1]} |f_n(x)-f(x)|= \sup\limits_{x \in [0,1]} f_n(x) - \inf\limits_{x \in [0,1]} f(x)$ for all $n \in \mathbb N$. Since $f_n \rightarrow f$ as $n \rightarrow \infty$ uniformly on $[0,1]$ so $M_n \rightarrow 0$ as $n \rightarrow \infty$. Hence we have $\sup\limits_{x \in [0,1]} f_n(x) \rightarrow \inf\limits_{x \in [0,1]} f(x)$ as $n \rightarrow \infty$. Which fails to meet my purpose.
What is wrong in my concept? Would anyone tell me please.
Thank you in advance.

Note that for $x\in I=[0,1]$,
$$f(x)=f(x)-f_n(x)+ f_n(x)\leq |f(x)-f_n(x)|+ f_n(x)\leq \|f-f_n\|+ \sup_{x\in I}f_n(x)$$
which implies that
$$\sup_{x\in I}f(x)\leq \|f-f_n\|+ \sup_{x\in I}f_n(x).$$
By symmetry,
$$\sup_{x\in I}f_n(x)\leq \|f-f_n\|+ \sup_{x\in I}f(x).$$
Finally, putting all together, we obtain
$$|\sup_{x\in I}f_n(x)-\sup_{x\in I}f(x)|\leq \|f-f_n\|.$$
