In Banach spaces is $X \cap Y = Z \Rightarrow \overline{{span} X} \cap \overline{{span} Y} = \overline{{span} Z}$ Let $V$ be a separable infinite dimensional Banach space over $\mathbb{C}$
Let $B \subset V$ be a subset of $V$ such that:
1) $B$ is linearly independent and closed
2) $\overline{\operatorname{span} B} = V$
3) For all $X$ and $Y$ closed subsets of $B$: $X \cap Y = \emptyset \Longrightarrow \overline{\operatorname{span} X} \cap \overline{\operatorname{span} Y} = \{0\}$
I would like to know if is it true that:
$X \cap Y = Z \Longrightarrow \overline{\operatorname{span} X} \cap \overline{\operatorname{span} Y} = \overline{\operatorname{span} Z}$
 A: I'm sorry for the last answer where I misunderstood the problem. Now I get it:
So I still think this isn't the case and suggest:$\require{begingroup}\begingroup\newcommand{\span}{\operatorname{span}}$
$V = l_{2}$.
$B = \{ \sum_{i=0}^{N} \left((\frac{1}{i} e_{2i + 1}) + e_{2i} \right) : i \in \mathbb{N}  \}  \cup \{ e_{2i} : i \in \mathbb{N}\}$. For simplicity lets define $f_N = \sum_{i=0}^{N} \left((\frac{1}{i} e_{2i + 1}) + e_{2i} \right)$ and $Z = \{ e_{2i} : i \in \mathbb{N}\}$.
$X = \{f_i : i  \equiv 0 \space \text{mod} \space 2 \} \cup Z $. and $Y = \{f_i: i \space \equiv 1 \space mod \space 2\} \cup Z$
first lets check that 1)$\overline{B} = B$ and that $B$ is linearly independent.
i) $\overline{B} = B$. To see this, first notice that $\langle e_{2n}, f_k\rangle =e_{2n}^{*}(b) \geq 1$ for each $k \in \mathbb{N}$ and each $n \leq \frac{k}{2}$. Assume $(b_l)_{l \in \mathbb{N}} \subseteq B$ is a non-constant sequence that converges to $v \in V$. Either $(b_l)_{l \in \mathbb{N}} $ has an infinite amount of $e_{2i}$ which contradicts that it converges, or it has an infinite amount of $f_k$ which would force $\langle e_{2n}, v\rangle \geq 1 $ for each $n \in \mathbb{N}$ which is also a contradiction.
ii) $B$ is linearly independent. Take 
$\sum_{i = 0}^N \lambda_i f_i + \sum_{j=0}^M \mu_je_{2j} = 0$.
By using that the only $\lambda$ (or $\mu$) that is affecting $e_{2N + 1}$ is $\lambda_N$, we get 
$\frac{\lambda_N}{N} = 0$. Then we have that $\lambda_N = 0$ so that gives us
$\sum_{i = 0}^{N - 1} \lambda_i f_i + \sum_{j=0}^M \mu_je_{2j} = 0$. Where we repeat the argument until we get that all $\lambda_i = 0$ for each $ i \leq N$, which leaves us with
$ \sum_{j=0}^M \mu_je_{2j} = 0$ which implies $\mu_j = 0$ for each $j \leq M$. So linear independence is established.
Now 2) $\overline{\span(B)} = V$,
To see this just notice:
$Z \subset \overline{\span(B)}$ and  $\sum_{i=0}^{N} \left((\frac{1}{i} e_{2i + 1}) + e_{2i} \right) \in \overline{\span(B)}$ for each $N \in \mathbb{N}$ implies
$\sum_{i=0}^{N} \frac{1}{i} e_{2i + 1} \in  \overline{\span(B)}$ for each $N \in \mathbb{N}$. which gives us
$\frac{1}{2N +1} e_{2N + 1} = \sum_{i=0}^{N} \frac{1}{i} e_{2i + 1} - \sum_{i=0}^{N-1} \frac{1}{i} e_{2i + 1} \in \overline{\span(B)}$ for each $N \in \mathbb{N}$. So that takes care of 2).
To see 3) we are gonna have to check some properties of $\overline{\span(B_1)}$for all $B_1 \subseteq B$. Basically, we are gonna assign every such $B_1$ a partition of the natural numbers in the following way:
Take $\{2n + 1: f_{n} \in B_1\}$, so the subset of all the "last odd numbers that each $f_n \in B_1$ knows, meaning the last odd number such that $e_i^{*}(f_n) \neq 0$", lets denote them in the ascending order as a posibly finite sequence (or even empty sequence)
$(n_1, n_2, \cdots, n_k , \dots)$
and make the partition, starting with $n_0 = 0$, $P_0 = [0, n_{1}]$ if $n_{1}$ exists and $P_0 = [0, \infty)$ if it doesn't exist, and then recursively $P_{k+1} = [n_k + 1, n_{k+1}]$ if $n_{k+1}$ exists and $P_{k+1} = [n_{k} +1 , \infty]$ if $n_{k+1}$ doesn't exists.
The property that we are looking for is the next one:
Let $x \in \overline{\span(B_1)}$ for each $k \in \mathbb{N}$ there exist a single $\lambda_k$ such that 
for each $l \in P_k$ 


*

*$e_l^*(x) = \frac{\lambda_k}{l}$ if $l \in 2\mathbb{N} + 1$, that is, if $l$ is odd.

*$e_l^*(x) = \lambda_k$ if $l \in 2 \mathbb{N}$ such that, $e_{l} \notin B_1$, that is, if $e_l \in (Z \setminus B_1 \cap Z )$


The proof goes by proving this property in $\span(B)$ and then using continuity of all the relevant $e_l^* = \langle e_l, \text{__} \rangle$ to establish it in the closed span. To see that it holds in $\span(B)$ it's enough to notice that every $B_1 \subseteq B$ has the property, and that everytime you take a linear combination 
$\gamma x + \mu f_n + \nu e_{2m}$ with $x$ having the property and $f_n, e_{2m} \in B_1$. Denote by $\{\lambda_{x,r}: r \in \mathbb{N}\}$, the lambdas for that specific $x$, and for their respective $\{P_r : r \in \mathbb{N}\}$. Then, $\nu$ doesn't change anything about the property because we purposely took those $2m$ away, and since $f_n = \sum_{i=0}^{N} \left((\frac{1}{i} e_{2i + 1}) + e_{2i} \right)$ such that $2N +1 = n_{k_0}$for some $k_0 \in \mathbb{N}$. Then, for every $P_r$ either assign $\lambda_r = \gamma \lambda_{x,r}  + \mu$ if $P_r \subseteq [0, n_{k_0}] $or  $ \lambda_r = \gamma \lambda_{x,r}$  if $P_r \subseteq [n_{k_0}+1, \infty]$. Notice that it cannot intersect both because of the definition of the $P_r$.
Therefore, $\overline{\span(B_1)}$ has such property.
So finally, take $B_1, B_2 \subseteq B$ such that $B_1 \cap B_2 = \emptyset$. We are gonna prove that $\overline{\span(B_1)} \cap \overline{\span(B_2)} = \{ 0\}$ using the property. Take $x \in \overline{\span(B_1)} \cap \overline{\span(B_2)} $, therefore $x$ has the property both for the partition given by $B_1$ and the partition given by $B_2$, lets call them $\{P_{B_1 , r} : r \in \mathbb{N}\}$ and $\{P_{B_2 , r} : r \in \mathbb{N}\}$, notice that the fact that $B_1 \cap B_2 = \emptyset$, gives us that 
a)$\max P_{B_1, r_1} \neq \max P_{B_2, r_2}$ for any $r_1, r_2 \in \mathbb{N}$. 
We are gonna prove that there exists a single $\lambda$ such that


*

*$e_l^*(x) = \frac{\lambda}{l}$ if $l \in 2\mathbb{N} + 1$, that is, if $l$ is odd.

*$e_l^*(x) = \lambda$ if $l \in 2 \mathbb{N}$ such that, $e_{l} \notin B_1$ or $e_{l} \notin B_2$. 


and this is done by first checking the bigger of $P_{B_1, 0}$ and $P_{B_2, 0}$, Suppose this is $P_{B_1, 0}$ and take that $\lambda_{B_1, 0}$, then whatever $P_{B_2, c_1}$ such that $\max P_{B_1, 0} \in P_{B_2, c_1}$, notice that since $x$ has this property with both subsets and $\max P_{B_1, 0} \in P_{B_2, c_1} \cap  P_{B_1, 0}$ is odd, this means $\lambda_{B_1, 0} = \lambda_{B_2, c_1}$. Because of $a)$ we have that $\max P_{B_2, c_1} >  \max P_{B_1, 0} $ so now choose $\max P_{B_2, c_1} \in P_{B_1, d_1}$ and then $P_{B_2, c_2}$ and recursively until you have 
$\mathbb{N} = \bigcup_{i=0}^{\infty} P_{B_1, c_i} \cup P_{B_2, d_i} $, such that $\lambda_{B_1, c_i} = \lambda_{B_2, d_j}$ for each $i,j \in \mathbb{N}$. Notice that this actually gives us that all $\lambda_{B_1, r} = \lambda_{B_2, r'}$ for $r, r' \in \mathbb{N}$ (simply check with any of the ones we know to be equal that it intersects).
Since, because they are disjoint, $(B \setminus B_1) \cup (B \setminus B_2) \supset B \supset Z $. We actually get that $e^{*}_n(x) = \lambda$ for each $n \in 2 \mathbb{N}$, each even $n$. That means that $\lambda = 0$, so $x = 0 $.
And now we just have to check that $X \cap Y = Z$ (that is clear) and
$ \overline{\span(X)} \cap \overline{\span(Y)} \neq \overline{\span{Z}}$, and that is easy to do because
$\sum_{i=0}^{\infty} (\frac{1}{i} e_{2i + 1}) \in \overline{\span(X)} \cap \overline{\span(Y)} $ and its clearly not in $\overline{\span(Z)}\endgroup$
