I want to better understand what characterize the vector spaces that have a Schauder basis.
Usually such a basis $ \{u_i \} \quad i\in \mathbb{N}$ is defined for a Banach space using the norm to define the convergence of the series that gives any vector $v$ of the space as:$$v=\sum_{i=0}^\infty a_i u_i $$
My question is if we can define such a countable basis also in a topological vector space that is not normable, so that it is not a Banach space. From Wikipedia it seems that this is possible using the convergence of sequences in the given topology, but I don't see how this can be concretely done. As an example: what is a Schauder basis for the space of distributions, that ( it seems to me) is a non-normable vector space ? Or for other similar spaces.
Added to clarify my problem.
What is strange for me is that, if a Schauder basis exists, such that $$v=\sum_{i=0}^\infty v_i u_i \qquad w=\sum_{i=0}^\infty w_i u_i $$ than we can define an inner product in the vector space as $$ \langle v,w\rangle=\sum_{i=0}^\infty v_i w_i $$ and, from this inner product, we have a norm... so the space is normable (contradiction or my mistake)
Given a topological vector space with a Schauder basis, you can attempt to define an inner product $\langle v,w\rangle=\sum_{i=0}^\infty v_i w_i$ as you have done. However, the sum defining this inner product may fail to converge. And even if it does converge, the inner product may induce a different topology than the given topology. So the mere existence of a Schauder basis certainly doesn't mean that the topology on your space must come from an inner product.
As a simple example of a Schauder basis in a non-normable space, consider the space $\ell^p$ for $1\leq p<\infty$ equipped with the weak topology (not the usual norm topology!). The sequences which have all entries $0$ except for a single $1$ form a Schauder basis.