Schauder bases are linearly independent. Indeed, let $(e_n)_{n=1}^\infty$ be a Schauder basis for a Banach space. If it were linearly dependent, then the zero vector would have two expansions contradicting uniqueness:
$$0= \sum_{k=1}^\infty 0\cdot e_k = \sum_{k=1}^\infty a_k e_k$$
where $a_k$ are eventually zero, but some of them are non-zero and add up to 0 when multiplied by $e_k$.
However, linear independence in not really the primary issue when dealing with Schauder bases.