$C([0,1])$ is a $C^*$ algebra of complex functionals. $A$ is a $C^*$ algebra. Hence $C([0,1],A)$, the continuous functions from $[0,1]$ to $A$ is also a $C^*$ algebra.
We construct its tensor product in the category of $C^*$ algebras. This is unique as $C([0,1])$ is nuclear.
There is a canonical $*$-homomorphism, $$C[0,1] \otimes A \rightarrow C([0,1],A)$$ $$(f \otimes a) (x) = f(x)a $$ How does one show this is in fact an isomorphism?
The map is $\pi:f\otimes a \longmapsto f a$. It's obvious that it is linear and multiplicative, and preserves adjoints, so it is a $*$-homomorphism. It is injective: if $\pi(\sum_j f_j\otimes a_j)=0$, we may choose the $a_j$ so that they are linearlity independent. Then $$ 0=\pi(\sum_j f_j\otimes a_j)=\sum_j f_j a_j. $$ Evaluating at any $t\in[0,1]$, we have $0=\sum_j f_j(t)a_j$, and the linear independence gives $f_j(t)=0$ for all $j$. As we can do this for any $t$, $f_j=0$ for all $j$, and so $\sum_jf_j\otimes a_j=0$. Thus $\pi$ is injective (in reality one also needs to check that the map is well-defined, see this answer for details).
It remains to show that $\pi$ is onto. Since it is an isometry, it is enough to show that its range is dense. There you use compactness of $[0,1]$ to show that the functions of the form $\sum_j f_j a_j$ are dense. Concretely, let $f\in C([0,1],A)$, Fix $\varepsilon>0$, then there exists a partition $0=t_0<t_1<\ldots<t_m=1$ such that $\|f(t_j)-f(t)\|<\varepsilon$ for $t\in[t_j,t_{j+1}]$. Let $a_j=f(t_j)$ for each $j$; let $g_j=1_{[t_j,t_{j+1}]}$, and let $f_j\in C[0,1]$ such that $\|f_j-g_j\|<\varepsilon/(m\|f\|)$. Then, with $k$ such that $t\in[t_k,t_{k+1}]$, \begin{align} \|f(t)-\sum_j f_j(t)a_j\| &\leq \|f(t)-\sum_j g_j(t)a_j\|+\|\sum_j(g_j(t)-f_j(t))a_j\|\\ &\leq \|f(t)-\sum_j g_j(t)a_j\|+\varepsilon\\ &=\|f(t)-f(t_k)\|+\varepsilon\\ \ \\ &<2\varepsilon. \end{align}