Proof about direct sum of vector spaces 
Let $V$, a vector space above $\mathbb{F}$ and $W_1,...,W_n \subseteq V$, such that $V = W_1 + ... + W_n$. Prove that $V = W_1 \oplus \ldots \oplus W_n \iff $ for every $I,J\subseteq \{1\ldots n\}$ such that $I\cap J = \emptyset$ and $I\cup J = \{1\ldots n\}$: $$\text{span}\{ W_i \}_{i\in I} \cap \text{span}\{ W_j \}_{j\in J} = \{0\}$$

I've read that for the generalized case: $V = W_1 \oplus \ldots \oplus W_n \iff $ for all $i$ $W_i \cap ( W_1 \ldots W_{i-1} + W_{i+1} + \ldots + W_n) = \emptyset$ so maybe it can come handy here.
I'd be glad for guidance.
Thanks in advance.
 A: Say $V = W_1 \oplus … \oplus W_n$.
If there was a pair $I, J \subseteq \{1, …, n\}$ such that $I ∩ J = ∅$ and $I ∪ J = \{1, …, n\}$, but 
$$\mathrm{span} \{W_i;~i ∈ I\} ∩ \mathrm{span} \{W_j;~j ∈ J\} ≠ \{0\},$$
you would be able to write $0 ∈ V$ in two different ways as a sum $0 = w_1 + … + w_n$ with $w_1 ∈ W_1, …, w_n ∈ W_n$. (There’s always a trivial one, but what’s the other one?)

Say the condition you have given holds.
Since $V = W_1 + … + W_n$, you can write any $v ∈ V$ as $v = w_1 + … + w_n$ for some $w_1 ∈ W_1, …, w_n ∈ W_n$. To show that this already is unique, it suffices to prove that $v = 0$ can be uniquely written as such a sum. (Why?) 
So write $0 = w_1 + … + w_n$ with $w_1 ∈ W_1, … , w_n ∈ W_n$. Then
\begin{align*}
w_1 &= -(w_2 + w_3 + … + w_n), &\text{so}& &w_1 ∈ &W_1 ∩ \mathrm{span} \{W_2, W_3, …, W_n\}\\
w_2 &= -(w_1 + w_3 + … + w_n), &\text{so}& &w_2 ∈ &W_2 ∩ \mathrm{span} \{W_1, W_3, …, W_n\}\\
&… &\text{so}& &…&\\
w_n &= -(w_1 + w_2 + … + w_{n-1}), &\text{so}& &w_{n-1} ∈ &W_{n-1} ∩ \mathrm{span} \{W_1, W_2, …, W_{n-1}\}.
\end{align*}
Therefore …
(Yeah, or put otherwise: If your condition holds, then trivially the weaker condition holds that you have given. So you can already conclude the sum is direct.)
