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.
 A: Hint: prove that $\|\cdot\|_{\infty}$ defined by $\|f\|_{\infty} = \sup_{x\in [0,1]} |f(x)|$ is a norm on $C([0,1])$ (the space of continuous functions on $[0,1]$). Then use the corollary that $|\|x\|-\|y\|| \le \|x-y\|$ of the triangular inequality for norms.
A: In addition to the statement of the result being wrong, your application of the given result isn't correct for other reasons; the result states that,
$$ \sup_{x \in S, y \in T} |x-y| = |\sup_{x \in S} x - \inf_{y \in T} y|. $$
So if you take $S_n = \{f_n(x) : x \in [0,1] \}$ and $T = \{f(x) : x \in [0,1]\},$ then we actually get,
$$ M_n = \sup_{s \in S_n, t \in T} |s-t| = \sup_{x \in [0,1]} \sup_{y \in [0,1]} |f_n(x) - f(y)| \neq \sup_{x \in [0,1]} |f_n(x)-f(x)| $$
in general. Indeed taking $f_n(x) = f(x) = x$ for all $n \in \mathbb N,$ we get $M_n = 1$ for all $n$ which does not vanish as $n \rightarrow 0.$
As others have already mentioned, their are easier ways of proving this result.
A: You have
$$ f(x) = f_n(x) + (f(x) -f_n(x)) \leq \sup_{y\in [0,1]} f_n(y) + \sup_{y\in [0,1]} (f(y) -f_n(y)). $$
Taking supremum over $x\in [0,1]$ gives
$$ \sup_{x\in [0,1] } f(x) \leq \sup_{y\in [0,1]} f_n(y) + \sup_{y\in [0,1]} (f(y) -f_n(y)). $$
Taking the $\liminf$ on both sides yields by uniform convergence
$$ \sup_{x\in [0,1] } f(x) \leq \liminf_{n\rightarrow \infty} \sup_{x\in [0,1]} f_n(x).$$
Now redo the argument.
$$ f_n(x) = f(x) + (f_n(x) -f(x)) \leq \sup_{y\in [0,1]} f(y) + \sup_{y\in [0,1]} (f_n(y) -f(y)). $$
Taking supremum over $x\in [0,1]$ gives
$$ \sup_{x\in [0,1] } f_n(x) \leq \sup_{y\in [0,1]} f(y) + \sup_{y\in [0,1]} (f_n(y) -f(y)). $$
Taking the $\limsup$ on both sides yields by uniform convergence
$$ \limsup_{n\rightarrow \infty}\sup_{x\in [0,1] } f_n(x) 
\leq  \sup_{x\in [0,1]} f(x).$$
Hence, we have
$$  \limsup_{n\rightarrow \infty}\sup_{x\in [0,1] } f_n(x) 
\leq  \sup_{x\in [0,1]} f(x) \leq \liminf_{n\rightarrow \infty}\sup_{x\in [0,1] } f_n(x).$$
Thus, we conclude
$$ \lim_{n\rightarrow \infty}  \sup_{x\in [0,1] } f_n(x)  
= \sup_{x\in [0,1] } f(x).$$
A: 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
$$|\sup_{x\in I}f_n(x)-\sup_{x\in I}f(x)|\leq \|f-f_n\|.$$
