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.

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 : $\parallel f(x)-\sum a_i(x)b_i \parallel < 3 \epsilon$.
