Show $F_b (\Omega, X)$ is a Banach space Let $\Omega$ be any non empty set and let $X$ be a Banach space over $\mathbb{C}$.
Let $F_b (\Omega,X)$ be a linear subspace of $F(\Omega, X)$ of all functions $f; \Omega \to X$ such that $$\{||f(x)|| : x \in \Omega\}$$ is bounded. 
The vector  space $F_b (\Omega , X)$ is a normed space with norm $$||f||_b =\sup \{||f(x)|| : x \in \Omega \}$$

I am trying to show $F_b (\Omega ,X)$ is a Banach space.

Proof: 
Let $(f_n)$ be a cauchy sequence in $F_b (\Omega,X)$ and let $\epsilon >0$.
Then there exists a $N \in \mathbb{N}$ such that $||f_n -f_m ||_b \leq \epsilon$ whenever $n ,m > N$.
For all $x \in \Omega$ we have $$||f_n(x)-f_m(x)|| \leq ||f_n -f_m||_b < \epsilon$$ whenever $n , m > N$. 
Hence $(f_n (x))$ is a Cauchy sequence in $X$.
Since $X$ is complete we may define a function $f : \Omega \to X$ by $$f(x)=\lim_{n \to \infty} f_n (x)$$

I understand the proof up to this point but not any further. 

The proof continues 
Since $$||f_n(x)-f_m(x)|| < \epsilon$$ for all $x \in \Omega$ whenever $n,m > N$ taking the limit $ m \to \infty$ yields $$||f_n(x) -f_m(x)|| \leq \epsilon$$ provided $n > N$

The answer says we have taken the limit but it still yields $||f_n(x)
 -f_m(x)|| \leq \epsilon$?
Why do we take the limit $m \to \infty$? And why not $n \to \infty$?

Hence $$f(x)|| \leq \epsilon + ||f_n(x)|| \leq \epsilon + ||f_n||_b$$ provided $n >N$ for all $x \in \Omega$.

Where does $f(x)|| \leq \epsilon + ||f_n(x)|| \leq \epsilon +
 ||f_n||_b$ come from?

Thus $f$ is bounded and belongs to $F_b (\Omega,X)$
and $\lim_{n \to \infty} f_n =f$
I understand the definition of a Banach space and I understand that to show completeness you have to show every Cauchy sequence is convergent in the space. 
So I understand what this proof is trying to do but it doesn't make sense.  

What I am looking for is a clearer, simpler and a more well explained proof to
  the question.

 A: To show completeness, take a Cauchy sequence $(f_n)_{n\in \mathbb{N}}\subseteq F_b(\Omega, X)$ and show, there exists $f\in F_b(\Omega, X)$ such that
$$ \lim_{n\rightarrow \infty} \Vert f - f_n \Vert_{b}=0.$$
As you already noted, one can construct a candidate for the limit of our sequence. Namely, we note that for fixed $x\in \Omega$ we have
$$ \Vert f_n(x) - f_m(x) \Vert_X \leq \sup_{y\in \Omega} \Vert f_n(x) - f_m(x) \Vert_X = \Vert f_n -f_m \Vert_b.$$
Hence, $(f_n(x))_{n\in \mathbb{N}}$ forms a Cauchy sequence in $X$ and so $f(x):=\lim_{n\rightarrow \infty} f_n(x)$ exists. This gives us our candidate.
First we need to check, whether $f\in F_b(\Omega, X)$. Indeed, let $x\in \Omega$, choose $n\in \mathbb{N}$ such that $\Vert f(x) - f_n(x) \Vert_X\leq 1$. Then
$$  \Vert f(x) \Vert_X \leq \Vert f(x)-f_n(x) \Vert_X + \Vert f_n(x) \Vert_X \leq 1+ \Vert f_n \Vert_b.$$
Hence,
$$ \Vert f \Vert_b \leq 1 + \sup_{n\in \mathbb{N}} \Vert f_n \Vert_b.$$
Show that the RHS is bounded (use the fact that $(f_n)_{n\in \mathbb{N}}$ is a Cauchy sequence). Now we are left to prove that $f_n \rightarrow f$. Let $\epsilon >0$. Fix $x\in \Omega$ and choose $N\in \mathbb{N}$ such that for all $n,m\geq N$ holds
$$ \Vert f_m - f_n \Vert_b \leq  \frac{\epsilon}{2}.$$
Furthermore, choose $m_x\geq N$ such that
$$ \Vert f(x) -f_{m_x}(x) \Vert_X < \frac{\epsilon}{2}.$$
For all $n\geq N$ we have then
$$\Vert f(x) -f_n(x) \Vert_X \leq \Vert f(x) -f_{m_x}(x) \Vert_X + \Vert f_{m_x}(x) -f_n(x) \Vert_X \leq \frac{\epsilon}{2} + \Vert f_n - f_{m_x} \Vert_b < \epsilon. $$ 
Thus for all $n \geq N$ we have
$$ \Vert f -f_n \Vert_b < \epsilon. $$
This establishs $f_n \rightarrow f$ and hence the completeness of $F_b(\Omega, X)$.
A: You have the following typos:


*

*the display after "yields" should read
$$
\|f_n(x)-f(x)\|<\epsilon
$$
because you've taken the limit for $m\to\infty$ with $n$ fixed and you've used the continuity of the norm in $X$.

*You're missing a $\|$ bracket before $f(x)\|$ elsewhere.
Now using the inequality in 1. and the triangle inequality (in the form $\|a\|\geq\|a+b\|-\|b\|$) you get
$$
\|f(x)\|<\epsilon+\|f_n(x)\|<\epsilon+\|f_n\|_b+\epsilon<B+\epsilon,
$$
where $B=\sup_n\|f_n\|_b$ which is finite by your assumption that $(f_n)_{n\geq1}$ is Cauchy.
Noting that the last display yields an inequality that does not depend on $n$ (look at the two extremes) and \emph{for any $\epsilon>0$}, it follows that
$$
\|f(x)\|\leq B
\text{ for all }
x\in\Omega,
$$
and thus $\|f\|_b\leq B$, whence $f\in F_b(\Omega,X)$.
