If $X$ is a Banach, then $X$ is indecomposable if $X = M \oplus N$ implies that $M$ or $N$ is finite dimensional.
An unconditional basis is a set $\{e_n\}$ such that every $x$ can be written uniquely as $$x = \sum_{n=1}^{\infty}a_ne_n$$ $(a_n = a_n(x))$ and the convergence is unconditional, i.e. it converges for every permutation (note this is implied by, but does not imply absolute convergence.)
As a consequence of the uniform boundedness theorem, the partial sum projections $s_n(x):= a_1 + ... + a_n$ are continuous.
Unconditional convergence of a sum $\sum_{n=0}^{\infty}x_n$ is equivalent to convergence of the sum along every increasing subsequence of integers.
Let $\mathbb{N} = \{n_k\} \sqcup \{n_j\}$. Does it follow that $$\sum_{n=1}^{\infty}a_ne_n \mapsto \sum_{j=1}^{\infty}a_{n_j}e_{n_j}$$ $$\sum_{n=1}^{\infty}a_ne_n \mapsto \sum_{k=1}^{\infty}a_{n_k}e_{n_k}$$
define bounded projections? I.e. are the spaces $\overline{span(e_{n_j})}$ and $\overline{span(e_{n_k})}$ complemented? I think this is why the resolution of the unconditional basic sequence problem implies the existence of a hereditarily indecomposable space (every closed subspace of $X$ is indecomposable), but I cannot tell if I'm missing something.
As already mentioned by David Mitra, these are bounded projections.
However, ranges of such projections are usually non-isomorphic. You can already find a counterexample in $\ell_1\oplus \ell_2$, but the most striking example is Tsirelson's space for which we have continuum many isomorphic types; this is Corollary VII.b.3 in