Direct sum equivalence definition on vector subspaces 
Let $W_i ,~(1 \leq i \leq s, s \geq 2)$ be vector subspaces of a
  vector space $V$ over a field $F$. Set $$W = W_1+W_2+\cdots+W_s$$ Then
  the following are equivalent.
i) We have $$W_1+\cdots+W_s = W_1 \oplus \cdots \oplus W_s$$ which means
   $W_1+\cdots+W_s$ is a direct sum of $W_i$.  
We say that the sum $W_1+W_2+\cdots+W_s$ is a direct sum of vector
  subspaces $W_1,\ldots,W_s$, if the intersection
  $$\left(\sum_{i=1}^{k-1}W_i\right) \cap W_k = \{0\},~~~(2 \leq \forall
 k \leq s)$$
ii) $$\left(\sum_{i \neq \ell}W_i\right) \cap W_\ell = \{0\},~~~(1
 \leq \forall \ell \leq s)$$

Now prove ii implies i. 
My argument is that if i can prove the following then we are done:
$$(W_1+W_3+\cdots+W_s) \cap W_2 = \{0\}  \Rightarrow W_1 \cap W_2 = \{0\}$$
$$(W_1+W_2+W_4+...+W_s) \cap W_3=\{0\}  \Rightarrow (W_1 +W_2) \cap W_3 = \{0\}$$
$$(W_1+W_2+W_3 + W_5+\cdots+W_s) \cap W_4 =\{0\}  \Rightarrow (W_1 +W_2+W_3) \cap W_4 = \{0\}$$
$$\vdots$$
$$(W_1+W_2+\cdots+W_{s-1}) \cap W_s =\{0\}  \Rightarrow (W_1 + W_2 + W_3 + \cdots + W_{s-1}) \cap W_s = \{0\}$$
First, I show that $(W_1+W_3+\cdots+W_s) \cap W_2 = \{0\}$ implies that $W_1 \cap W_2 = \{0\}$. Intuitively speaking, it is very obvious. Because our hypothesis means you take any vectors from $W_1,W_3,\ldots,W_s$ and add them up in any way, and you find that the intersection with $W_2$ is still $0$, then obviously if you just take any vector from $W_1$ and "intersect" with $W_2$, you cannot find a common vector between them except $0$ as well. Or else it will be a contraction, like if there is a common vector ${b}$ from $W_1 \cap W_2$, then that means $(W_1+\cdots+W_s) \cap W_2$ cannot be equals to $0$ already. Because let us take $b$ from $W_1$ and $0 \in W_3,W_4,\ldots,W_s$, then add them up it will be equals to $b$ : $$b + 0+0+\cdots+0 = b$$ but since $b$ is also in $W_2$, it follows that there does exist a element $b \in (W_1+W_3+\cdots+W_s) \cap W_2$.
That said, do note that $W_i$ belong the same vector space, so i can easily assume they all have the same zero vector. That is why i can take the same $0 \in W_3,W_4,\ldots,W_s$.
Now we prove $$(W_1+W_2+W_4+\cdots+W_s) \cap W_3 = \{0\} \Rightarrow (W_1+W_2) \cap W_3 = \{0\}$$ Let us assume a contradiction that $(W_1+W_2) \cap W_3 \neq \{0\}$, then there exists $x \in (W_1+W_2) \cap W_3$ which means $x \in W_1+W_2$ and $x \in W_3$. 
So let us take $x \in W_1+W_2$, then $x = x+0+0+0 +\cdots+0 \in W_1+W_2+W_4+\cdots+W_s$, but then $x \in W_3$ by our assumption, then it contradicts that $(W_1+W_2+W_4+\cdots+W_s) \cap W_3 = \{0\}$.
And we can keep going on until all is shown.
Can anyone verify if my line of proof makes sense or is it correct? How do i make it more rigourous? **

P.S: Besides giving your own answer, i do hope people can verify my
  line of logic so that i can know where i gone wrong or right. THanks!

**
 A: I think you're on the right track, but let's make this a bit more rigorous by covering all of the conditional statements you're trying to prove. For each $\ell,$ $2\leq \ell\leq s,$ we want to prove $$\left(\sum_{k\neq\ell}W_{k}\right)\cap W_{\ell}=\{0\}\Rightarrow \left(\sum_{k=1}^{\ell-1}W_{k}\right)\cap W_{\ell}=\{0\}.$$ We could do this by contradiction, but let's try for a direct proof. We may write $\sum_{k\neq \ell}W_{k}=\left(\sum_{k=1}^{\ell-1}W_{k}\right)+\left(\sum_{k=\ell+1}^{s}W_{k}\right),$ which makes it clear that $\sum_{k=1}^{\ell-1}W_{k}\subseteq \sum_{k\neq \ell}W_{k}$ (as you mentioned, for $x\in \sum_{k=1}^{\ell-1}W_{k},$ $x+0$ clearly respects the decomposition we just proved for $\sum_{k\neq\ell}W_{k}$). Let $x\in \left(\sum_{k=1}^{\ell-1}W_{k}\right)\cap W_{\ell}.$ Then $x\in \sum_{k=1}^{\ell-1}W_{k},$ so $x\in \sum_{k\neq \ell}W_{k},$ and $x\in W_{\ell},$ which means $x\in \left(\sum_{k\neq \ell}W_{k}\right)\cap W_{\ell}=\{0\},$ which means $x=0.$ Since this is true for any element of $\left(\sum_{k=1}^{\ell-1}W_{k}\right)\cap W_{\ell},$ we see that this set must be exactly $\{0\}.$ Since this holds for each of the values of $\ell$ of interest, $2\leq \ell\leq s-1$, this completes the proof (note that when $\ell=s,$ the antecedent and consequent of the conditional we're trying to prove are identical, so there is nothing to prove).
