The normed vector space of continuous function is complete First of all, let $(f_n)$ be a Cauchy sequence in $B(x)$ which is the vector space of bounded functions $f\colon X \to \mathbb R$ equipped with the norm $\|f\| = \sup|f(x)|$.
Note that $|f_n(x)-f_m(x)| \leq \|f_n-f_m\|$ which tends to $0$, therefore $f_n$ is a Cauchy sequence in $\Re$, hence we can define a function
$f\colon X \to \mathbb R$ s.t $f(x)= \lim f_n(x)$.
Now we have to show that $f$ is in $B(X)$, $f_n \to f$ in $B(X)$ and that $f$ is continuous. 
I am OK with the proofs for the first $2$. For the last one - do I have to show that it is continuous wrt the norm, or with respect to the usual distance metric?  
Edit: $(X,d)$ is a metric space
 A: The vector space $B(X)$ of bounded functions $f\colon X\to\mathbb{R}$ makes sense with no further structure on $X$.
What you have to show is that, given a Cauchy sequence $(f_n)$ in $B(X)$ there exists a bounded function $f\in B(X)$ such that
$$
\lim_{n\to\infty}f_n=f
$$
First of all, we can identify a function that should be the limit (provided it is bounded). Indeed, if $x\in X$, the sequence $(f_n(x))$ in $\mathbb{R}$ is Cauchy (easy proof), so we can define
$$
f(x)=\lim_{n\to\infty}f_n(x)
$$
Note that if the sequence converges to some function $g$ in $B(X)$, then it must be $\lim_{n\to\infty}f_n(x)=g(x)$, for every $x\in X$. Thus only the “pointwise limit” above can work.
Our tasks now are


*

*to prove that $f$ is bounded, and that

*$\lim\limits_{n\to\infty}f_n=f$.


Let's try point 1. For every $\varepsilon>0$, there exists $N_\varepsilon$ such that, for $m,n>N$,
$$
\|f_n-f_m\|<\varepsilon
$$
This implies, for all $n>N$, fixing any $m>N$,
$$
f_m(x)-\varepsilon<f_n(x)<f_m(x)+\varepsilon
\qquad\text{(for all $x\in X$)}
$$
Passing to the limit for $n\to\infty$, we get
$$
f_m(x)-\varepsilon<f(x)<f_m(x)+\varepsilon \tag{*}
$$
so $f$ is indeed bounded (fill in the details).
On the other hand, (*) holds for every $m>N_\varepsilon$,
so
$$
|f(x)-f_m(x)|<\varepsilon
$$
for every $x\in X$ and so $\|f-f_m\|<\varepsilon$, for every $m>N_\varepsilon$.

Now, suppose all $f_n$ are continuous; let $x\in X$ and fix $\varepsilon>0$. There is $N$ such that, for $n>N$, $\|f-f_n\|<\varepsilon/3$, for every $n>N$. Choose one $n>N$; then there is $\delta>0$ such that $d(x,y)<\delta$ implies
$$
|f_n(x)-f_n(y)|<\varepsilon/3
$$
Then, if $d(x,y)<\delta$, we have
$$
|f(x)-f(y)|\le
|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|<\varepsilon
$$
A: To prove the completeness, we need to prove for any Cauchy sequence $f_1,f_2,\dotsm \in E$  satisfying that
$\forall \epsilon >0$, there exists a number $n_0$ such that for all $m,n\ge n_0$
$$
||f_m-f_n||<\epsilon\\
\max_{x\in[a,b]} |f_m(x)-f_n(x)|<\epsilon
$$
that implies
$$
|f_m(x)-f_n(x)|<\epsilon \quad \forall n,m\ge n_0,\forall x\in[a,b]
$$
that means $(f_n(x))$ is Cauchy sequence $\forall x\in[a,b]$.
We denote 
$$
f(x)=\lim_{n\rightarrow \infty } f_n(x),\forall x\in[a,b]
$$
Let $n\rightarrow \infty $ , we have
$$
|f_m(x)-f(x)|<\epsilon,\forall m\ge n_0 ,\forall x\in[a,b]\\
\max_{x\in[a,b]}|f_m(x)-f(x)|<\epsilon \\
||f_m-f|<\epsilon
$$
Then we prove $f\in E$ 
Let $\forall x_0\in[a,b]$. Since $f_{n_0}$ is continuous on $[a,b]$, there exists $\delta >0$ such that $|f_{n_0}(x_0)-f_{n_0}(y)|<\epsilon$ for every $y\in[a,b]$ and $|x_0-y|<\delta$
$$
|f(x_0)-f(y)|= |f(x_0)-f_{n_0}(x_0)+f_{n_0}(x_0)-f_{n_0}(y)+f_{n_0}(y)-f(y)|\\
\le |f(x_0)-f_{n_0}(x_0)|+|f_{n_0}(x_0)-f_{n_0}(y)|+|f_{n_0}(y)-f(y)|\\
< 3\epsilon
$$
whenever $|x_0-y|<\delta$, that means $f$ is continuous.
Q.E.D.
