My question is how to guarantee the space with the basis defined complete?
More specifically, let discrete group $G$ act freely on cw-complex $Y$ by permuting its cells, $K_iY$ the cellular chain group of $Y$, We write $l^2G$ for the Hilbert space of square summable functions $f:G\to \mathbb{R}$. Define $C_i(Y,G):=l^2G\otimes_G K_iY$. Then we define the Hilbert space as following:
Here how to see $C_iY$ is complete?
If you define a Hilbert space by providing a (countable or uncountable) set of orthonormal vectors then the Hilbert space itself is the completion of the (pre) Hilbert space defined by the finite linear combinations of the basis (with the scalar product defined via the orthonormality relations of the basis).
This is complete by definition.
To be more explicit, in response to a comment: if you define a set $\{\phi_i\}_{i\in I}$ for some index set $I$ to be an orthonormal base, then the scalar product on the vector space of finite linear combinations is defined by letting $$\langle \sum_{i=0}^n a_{k_i} \phi_{k_i}, \sum_{j=0}^m b_{l_j} \phi_{l_j}\rangle := \sum_{i=1}^n\sum_{j=1}^m a_{k_i} b_{l_j} \langle\phi_{k_i}, \phi_{l_j}\rangle = \sum_{i=1}^n\sum_{j=1}^m a_{k_i} b_{l_j} \delta_{k_i l_j}$$ where the last equality is just the statement that the $\phi_i$ form an orthonormal base.