Tensor product space with projective norm is incomplete

The projective tensor product $\ell_1\widehat{\otimes}X$ is naturally isometrically isomorphic to the $\ell_1$-sum of countably many copies of $X$. The uncompleted tensor product $\ell_1 \odot X$ is then the linear span of elements of the form $(\xi_n x)$, where $(\xi_n)$ is in $\ell_1$ under this identification, which is hardly complete as there exist infinite convergent series. For example, take a linearly independent sequence $(x_n)_{n=1}^\infty$ of unit vectors in $X$ and consider $(n^{-2}x_n)_{n=1}^\infty$; it does not belong to (the image of) $\ell_1\odot X$.

In general the proof goes along the same lines -- it uses the possibility of (non-unique) representation of elements of the projective tensor product as infinite series of simple tensors. You then have to show that if $X$ and $Y$ are infinite-dimensional then there is an infinite series that cannot be truncated to a finite one.

It wasn't clear to me, immediately, how to do the general case, but I think you can use Biorthogonal Systems. Indeed, let $(x_n)\subseteq X$ and $(f_n)\subseteq X^*$ satisfy that $f_n(x_m) = \delta_{n,m}$. Let $(y_n) \subseteq Y$ be a linearly independent sequence. Set $$\tau = \sum_{n=1}^\infty \|x_n\|^{-1} \|y_n\|^{-1} 2^{-n} x_n\otimes y_n \in X \widehat\otimes Y. $$ Suppose, towards a contradiction, that $\tau \in X\otimes Y$. Now, $\tau$ induces a bounded linear map $T:X^*\rightarrow Y$, and by our assumption, $T$ is finite rank. However, clearly $T(f_n) = \|x_n\|^{-1} \|y_n\|^{-1} 2^{-n} y_n$ and so $(T(f_n))$ is a linearly independent sequence, contradicting $T$ being finite rank.

To construct $(x_n), (f_n)$ we can follow an old argument due to Markushevich. The argument is not hard, but is a touch long to type out here. The best link I could find was: Biorthogonal Systems in Banach Spaces (Google books). Search for "M-basis".