showing a space with sup norm is complete show that $(c_0,||\cdot||_\infty)$ be the space of real valued sequences converging to $0$ with the supremum norm is complete 
Okay so I know that the normed space is complete if any Cauchy sequence in my normed space converges to a value which is inside my normed space.
but how would I use this to show that my normed space is complete? is it enough to find one example inside my normed space and show that it is Cauchy and that it converges to $0$??
 A: No that won't be enough. 
You have to proceed that way:
Let's $(U_n)$ be a Cauchy sequence of $c_0$ (then $U_n$ is a sequence of sequence).
Thus : $\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p,q \geq N \parallel U_p-U_q\parallel _{\infty}<\epsilon$
Then you have to prove that $(U_n)$ converges to a certain $(U) \in C_0$.
EDIT :
We have an equivalence between (in fact this is more the definition of completeness for a metric space) :


*

*$(c_0, \parallel \cdot \parallel )$ is complete,

*Every Cauchy sequence of $c_0$ converges (for the sup norm).


So if you take any Cauchy sequence $(U_n)$ of $c_0$ and proove that this sequence converges to $U$ in $c_0$ for the sup norm, then you have won.
Answer :
First we are going to prove that the sequence (of sequences) $(U_p)$ converges to a certain $(U)$. Let's get back to the definition of $(U_p)$ being a Cauchy sequence :
$$\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p,q \geq N \parallel U_p-U_q\parallel _{\infty}<\epsilon$$
By the definition of the sup norm, this implies $$\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p,q \geq N,\forall n \in \mathbb{N}, |U_p^n-U_q^n|<\epsilon$$
Where $U_p^n$ is the $n$-th element of $U_p$ and the norm being the usual distance on $\mathbb{R}$. But $\mathbb{R}$ is complete, and $(U_p^n)_p$ is a real Cauchy sequence, so : $$\forall n \in \mathbb{N},U_p^n \underset{p \rightarrow \infty}\longrightarrow U^n$$
where $U^n$ is the $n$-th element of the sequence $(U)$. So now we have proven that $(U_p)$ converges to a sequence $(U)$, now let's prove that $(U) \in c_0$.
Because of the convergence of $(U_p)$, we have :
$$\forall \epsilon >0, \exists N \in \mathbb{N}, \forall p \geq N,\forall n \in \mathbb{N}, | U_p^n-U^n |<\epsilon $$
But, $$\forall p,\forall \epsilon' >0 ,\exists N' \in \mathbb{N}, \forall n>N', |U_p^n-0|<\epsilon'$$ because $\forall p, (U_p) \in c_0$. Let's combine the two.Thus,
$$\forall \epsilon,\epsilon' >0, \exists N,N' \in \mathbb{N}, \forall p \geq N,\forall n >N', |U^n-U_p^n|<\epsilon \text{ and } |U_p^n|<\epsilon'$$
By elementary manipulations on thoose inequalities, we get to $|U^n|<\epsilon + \epsilon'$. So the proof is complete.
