If $K$ is compact, then $C(K,\mathbb{R}^n)$ is a Banach space under the norm $\|f\|=\sup_{x\in K} \|f(x)\|$ 
Let $K$ be a topological space that is compact. Show that the space $C(K,\mathbb{R}^n)$ of all the continuous functions $f:K\to\mathbb{R}^n$ is a Banach space with the norm $\|f\|=\sup_{x\in K} \|f(x)\|$.

Below is my attempt:


*

*$\|f\|=\sup_{x\in K} \|f(x)\|$ is a norm:
i. Suppose that $\|f\|=\sup_{x\in K} \|f(x)\|= 0$; then $f=0$. Conversely, suppose that $f=0$. Then $0=\|f\|=\sup_{x\in K} \|f(x)\|$.
ii. $\|\alpha f\|=\sup_{x\in K} \|\alpha f(x)\|=\sup_{x\in K} |\alpha|\|f(x)\|=|\alpha|\sup_{x\in K} \|f(x)\|=|\alpha|\|f\|$
iii. $\|f+g\|= \sup_{x\in K} \|f(x)+g(x)\|\leq \sup_{x\in K} \|f(x)\| + \|g(x)\|= \sup_{x\in K} \|f(x)\| + \sup_{x\in K} g(x)\|$

*$C(K,\mathbb{R}^n)$ is complete (my doubt is here)
i. For all $\epsilon>0$, $\exists\, n_{0}$ such that if $m,n>n_{0}$ then $d(f_m,f_n)<\epsilon$.
We know that $|f_m-f_n|\leq \|f_m -f_n\|= \sup_{x\in K} \|f_m(x)-f_n(x)\|< \epsilon $ but how can I prove the last inequality?
ii. How to prove that $f_n$ converges to an $f$ in $C(K,\mathbb{R}^n)$?
 A: Neon's solution has some loose ends. So here I will use a very standard argument in real analysis.
Let $\{f_k\}\subset C(K,\mathbb{R}^n)$ be a Cauchy sequence under $\|\cdot\|$.
Step 1:  First to prove is the uniform boundedness of this sequence. 
For a fixed $1>\epsilon>0$, we can find an $N\in \mathbb{N}$ such that for $n,m>N$, $\|f_n-f_m\|<\epsilon$. Hence for any $n>N$, 
$$
\|f_n\| \leq \|f_{N+1}\| + \|f_n - f_{N+1}\|<\|f_{N+1}\|+1.
$$
Therefore there is a bound for any member in this sequence:
$$
\text{For all }k : \, \|f_k\| \leq M<\infty,$$
where $M$ is the bigger one in $\max\limits_{1\leq i\leq N}\|f_i\|$ and $\|f_{N+1}\|+1$.
Step 2: Find the pointwise limit $f$.
Define the function $f$ be
$$
f(x) := \lim_{n\to \infty} f_n(x).
$$
This is a well-defined pointwisely bounded function thanks to the facts that (a) $\{f_k\}$ is uniformly bounded under the supremum norm, (b)$\{f_k(x)\}$ is henceforth Cauchy in $|\cdot|$ for any fixed $x\in K$(my $|\cdot|$ here is $\|\cdot\|$ in your definition for $\|f_n(x)\|$, I just want to distinguish it from the supremum norm), (c)$\mathbb{R}^n$ is complete under the standard Euclidean norm $|\cdot|$. 
Step 3: Prove $f_k\to f$ under $\|\cdot\|$.
Using the definition of the sup norm leads to:
$$
\|f_k - f\|:= \sup_{x\in K}|f_k(x) - f(x)|
\\
= \sup_{x\in K} \lim_{n\to \infty}|f_k(x) - f_n(x)|
\\
\leq \liminf_{n\to \infty}\, \sup_{x\in K}|f_k(x) - f_n(x)|
\\
\leq \lim_{n\to \infty} \|f_k -f_n\| \to 0 \text{ as }k\to \infty.
$$
The second last inequality can be argued using:
$$
|g_n(x)|\leq \sup_{x\in K} |g_n(x)|\implies \liminf_{n\to \infty} |g_n(x)|\leq \liminf_{n\to \infty} \,\sup_{x\in K} |g_n(x)|
\\
\implies \sup_{x\in K}\lim_{n\to \infty} |g_n(x)|= 
\sup_{x\in K}\liminf_{n\to \infty} |g_n(x)|\leq \liminf_{n\to \infty} \,\sup_{x\in K} |g_n(x)|.
$$
Hence we have the convergence under the sup norm, or rather to say uniform convergence.
Step 4: Prove this $f\in C(K,\mathbb{R}^n)$.
This is fairly straightforward using the $\epsilon/3+\epsilon/3+\epsilon/3$ argument together with the uniform convergence we get from step 3:
$$
|f(x)-f(y)| \leq |f_k(x)-f(x)|+|f_k(y)-f(y)| + |f_k(x)-f_k(y)|.
$$
A: I understand $f_n$ is a Cauchy sequence in $C(K,R^n)$. We need to get a $f\in C(K,R^n)$ such that $f_n\rightarrow f$ in $C(K,R^n)$. To find such $f$, Note that  $||f_n(x)-f_m(x)||=||(f_n-f_m)(x)||\le Sup_{x\in K}||(f_n-f_m)(x)||=||f_n-f_m||$ as if $f_n\in C(K,R^n)$ and $f_m\in C(K,R^n)$ then $f_n-f_m\in C(K,R^n)$ (defined as usual pointwise way). In fact, $ C(K,R^n)$ is a real vector space. So, if $f_n$ is a Cauchy sequence, then $f_n(x)$ (in $R^n$)is a Cauchy sequence $\forall x\in K$. But $R^n$ is complete, so, $f_n(x)$ converges to some element in $R^n$. Now, the candidate function $f$ can be defined as $f(x)=\lim_n f_n(x)$. Now we can to prove $f\in C(K,R^n)$ and $f_n\rightarrow f$ in $C(K,R^n)$
A: Continuity at $x\in K$ for $f$ can be seen as  $|f(y)-f(x)|\le |f(y)-f_n(y)|+|f_n(y)-f_n(x)|+|f_n(x)-f(x)|$ and the fact $f_n$ is a Cauchy sequence in $C(K,R^n)$. Also, $f_n \rightarrow f$ can be easily seen.
