I am trying to prove something that seemed simple to me at first sight but apparently it is giving me a hard time. Here is the same question on MSE.

Let $E\subset A$ be a finite dimensional operator system with Hamel basis $\{x_1,\dots,x_n\}$ and let $B$ be an arbitrary $C^*$-algebra. I am trying to show that, if $\|\cdot\|$ is the minimal norm on $E\odot B$ (i.e. the spatial norm) and $E\otimes B:=\overline{E\odot B}^{\|\cdot\|}$, then every element of $E\otimes B$ is written uniquely as $\sum_{i=1}^nx_i\otimes b_i$. This is of course equivalent to showing that $E\odot B$ is complete with the minimal norm.

It is trivial to verify the claim for the elements of the algebraic tensor product $E\odot B$, by the following lemma:

If $X,Y$ are vector spaces and $\{a_1,\dots,a_n\}\subset X$ are linearly independent, then $$\sum_{i=1}^na_i\otimes b_i=0\implies b_1=\dots=b_n=0.$$

This lemma also takes care of uniqueness for us. But what about elements of $E\otimes B$ in general? Of course, the ideal thing would be to establish an inequality of the form $\|x_j\otimes b_j\|\leq C\|\sum_{i=1}^nx_i\otimes b_i\|$, but how?

I tried to consider $B\oplus\dots\oplus B\to (E\odot B,\|\cdot\|)$, $(b_1,\dots,b_n)\mapsto\sum_{i=1}^nx_i\otimes b_i$ and observed that this is bounded and bijective. The annoying thing: if I knew that $E\odot B$ is complete (which is what I want to show), then I could apply the open mapping theorem to deduce that the inverse is bounded, hence obtain the desired estimate.

I also tried an induction argument on $\dim(E)$ but it didn't work out.