It is well known that every subspace $Y$ of a separable Banach space $X$ is quasi-complemented.We denote its quasi-complement by $Z$. This is a classical result due to F. J. Murray and G. Mackey.The proof is based on the existence of $\{(y_n, f_n)_{n} \cup (z_n, g_n)_n \}$ which is a biorthogonal system of $X$ with $(y_n)_n, (z_n)_n $ generate the subspaces $Y,Z$ respectively and $(\bigcap_n \operatorname{Ker} f_n) \cap(\bigcap_n \operatorname{Ker} g_n) = \{ 0 \}, $
Theorem : Let $ X $ be a separable Banach space, $Y$ be an infinite dimensional subspace with infinite dimension quasi-complement $Z$. Choose $(y_k,f_k)_{k\in N}$, $(z_k, g_k)_{k\in N}$ the biorthogonal system associated to $Y, Z$.
For a given $n\in N$ choose $(a_i, b_i)_{i=1}^n$ in the unit sphere of $R^2$ with $a_ib_j - b_ia_j \neq 0$ and for each $i= 1\dots n$ set $M_i$ be the closed linear span of the sequence $\{a_iy_k+b_iz_k\}_k$.
Then for every $i \neq j$ the subspaces $M_i , M_j $ satisfy
$M_i \cap M_j = \{0\}$ and $M_i + M_j$ dense in $X$.
Proof : Fix $i \neq j$ and consider the following biorthogonal system
$\{ (a_iy_k + b_iz_k, b_jf_k - a_jg_k)_k\cup (a_jy_k + b_jz_k, b_if_k - a_ig_k)_k$.
This corresponds to the pair $M_i M_j$. It is easy to see that $ (y_k)_k \cup (z_k)_k \subset M_i + M_j $ hence $M_i + M_j$ is dense in $X$.
For similar reasons
$\langle \{ b_jf_k-a_jg_k \}_k \cup \{ b_if_k-a_ig_k \}_k \rangle = \langle \{f_k\}_k \cup \{g_k\}_k \rangle $
which is $ w^* $ dense in $X^*$.Hence
$(\bigcap_k \operatorname{Ker} b_jf_k- a_jg_k) \cap(\bigcap_k \operatorname{Ker} b_if_k-a_ig_k) = \{ 0 \}$
which yields that $M_i \cap M_j = \{0\}$.
Remark 1 : Clearly there exists a sequence $\{M_i\}_{i\in N}$ satisfying the conclusion of the theorem.
Remark 2 : The theorem remains valid for non separable reflexive Banach spaces and more generally for WCG spaces.
Indeed let $X$ be a WCG space. From Amir Lindenstrauss theorem there exists a family $(P_{\alpha})_{\alpha\in A}$ mutually orthogonal bounded projections such that for each $\alpha\in A$ we have that $P_{\alpha}(X)$ is separable and $\langle \bigcup_{\alpha}P_{\alpha}(X) \rangle$ is dense in $X$. For a fixed $ n\in N$ and every $ \alpha\in A$ choose a family $(M_i^{\alpha})_{i=1}^n$ of closed subspaces of $P_{\alpha}(X)$ which are pairwise quasi-complemented in $P_{\alpha}(X)$.
For $ i= 1\dots n$ set $M_i = \overline{\langle\bigcup_{\alpha}M_i^{\alpha}\rangle}$. The family $(M_i)_{i=1}^n$ has the desided property.Indeed it is clear that for $i \neq j$ $ M_i + M_j $ is dense in $X$. To see that $M_i \cap M_j =\{ 0 \}$ assume on the contrary that there exists $x \in M_i \cap M_j$ with $x\neq 0$.Then there exists $ \alpha\in A$ such that $P_{\alpha}(x) \neq 0$.Since $P_{\alpha}(M_q) = M_q^{\alpha}$ for all $q= 1\dots n$ we derive a contradiction.
I also found interesting the question, stated by Janko Bracic, if the spaces $(M_i)_i$ could be selected to be pairwise isomorphic. It has an affirmative answer and is explained below.
Remark 3 : Assume that for all $k\in N$ $\|f_k\| = \|g_k\|=1$ and $f_k(y_k) = g_k(z_k) =1$.For $i= 1 \dots n $ and $k\in N$ choose $a_i^k, b_i^k$ and $\epsilon_k > 0$ such that the following hold.
For $i \neq j$ $\frac{1}{|a_i^k|}(|a_i^k - a_j^k|\|y_k\| + |b_i^k - b_j^k|\|z_k\|) < \epsilon_k$ and $\sum_k \epsilon_k < \frac{1}{2}$.
For every $k\in N$ and $i\neq j$ we have $a_i^k, a_j^k \neq0$ and $ a_i^k b_j^k- b_i^ka_j^k\neq 0$.
Define
$M_i = \overline { \langle (a_i^ky_k + b_i^kz_k)_k \rangle}$
It follows easily that the new family $(M_i)_{i=1}^n$ satisfies the conclusion of the theorem.We claim that the spaces are pairwise isomorphic. Indeed let $ y_i = \sum_{k=1}^l c_k(a_i^ky_k + b_i^kz_k)$ be a normalized vector in $M_i$. Then $|f_k (y_i)| =|c_ka_i^k|\leq 1$ hence $|c_k|\leq \frac{1}{|a_i^k|}$.
This yields that $\|y_i -y_j\|\leq \frac{1}{2}$ which shows that $M_i, M_j$ are isomorphic.
Remark 4 : There is something which I could not understand. If $M_i, M_j$ are spaces as in Remark 3 then $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$. Hence $M_i, M_j$ are not a complemented pair. But if the initial subspaces $Y, Z$ are a complemented pair then $M_i + M_j = Y+Z= X$ hence $M_i, M_j$ is a complemented pair. Any help?
Remark 5 : There is no problem concerning the proof of Remark 3. My false was that I consider that $M_i + M_j = Y+Z$. What actually we have is that $\langle (y_k)_k \cup (z_k)_k \rangle \subset M_i +M_j$ which does not imply that $\overline {\langle (y_k)_k \rangle} + \overline{ \langle (z_k)_k \rangle} \subset M_i +M_j$.
Therefore Remark 3 yields that independently to what is the initial quasi-complemented pair $Y,Z$( complemented or not) the $(M_i)_i$ resulting from Remark 3, since $\operatorname{dist} (B_{M_i}, B_{M_j}) = 0$, are pairwise quasi-complemented but not complemented. Hence Remark 3 answers completely the posed question for separable and WCG spaces.
