Show that $C(K,E)$ is separable This is an exercise of the book Analysis III of Amann and Escher:

Let $E$ a separable Banach space over $\Bbb K$ (being $\Bbb K=\Bbb R$ or $\Bbb K=\Bbb C$) and $K$ a compact metric space. Show that $C(K,E)$ is separable.
HINT: choose $A\subset C(K,\Bbb K)$ and $B\subset E$ dense and countable and consider $$H:=\left\{\sum_{k=0}^m a_j(x)b_j:m\in\Bbb N,\,(a_j\times b_j)\in A\times B,\, j=1,\ldots ,m\right\}$$

I already solved this exercise but I dont used the hint, so Im curious about what the hint is trying to point.
I know that $C(K,\Bbb K)$ is separable, so I guess that the hint is about taking $A$ countable and dense in $C(K,\Bbb K)$. But assuming that I dont know how to show that $H$ is dense in $C(K,E)$.
Someone figure out how to solve the exercise using the hint? Thank you.
P.D.: I noticed now that my other way to try to solve this exercise was wrong, so this exercise is unsolved at this moment.
 A: Maybe it is not very clear but it could give you an idea for the solution.
Take $f\in C(K,E)$ and $\epsilon >0$. Since $K$ is compact, $f(K)$ is compact so it is covered by a finite number of balls $B(b_i,\epsilon)$, $i=1,...,n$, where $b_i \in B$. Also, $f$ is uniformly continuous : there exists $\alpha >0$ such that $\forall x,y \in K, d(x,y) < \alpha \Rightarrow \parallel f(x)-f(y) \parallel < \epsilon$. 
Let $U_i=f^{-1}(B(b_i,\epsilon))$. We can define a continuous function $\phi_i$ on $K$ such that $\phi_i$ is $1$ on $U_i$ and is $0$ for $x \in K$ such that $d(x,U_i) \geqslant \alpha$. Choose $a_i \in A$ such that $\parallel \frac{\phi_i}{\phi_1+ \cdots + \phi_n}-a_i \parallel < \epsilon/(n \max(\parallel b_i \parallel)$.
Then for $x \in K$, $f(x)=\sum \frac{\phi_i}{\phi_1 + \cdots +\phi_n}(x)f(x)$, so $f(x)- \sum a_i(x)b_i= \sum \frac{\phi_i}{\phi_1 + \cdots +\phi_n}(x)(f(x)-b_i) + \sum (\frac{\phi_i}{\phi_1 + \cdots +\phi_n}(x)-a_i(x))b_i$. But $\vert \phi_i(x)(f(x)-b_i)\vert <2 \epsilon$ and $\vert \sum \frac{\phi_i}{\phi_1 + \cdots +\phi_n}(x)-a_i(x))b_i\vert < \epsilon$. Finally : $\parallel f(x)-\sum a_i(x)b_i \parallel < 3 \epsilon$.
