How does the equivalence of $(1)$ and $(3)$ hold? I am studying linear algebra and I viewed an online NPTEL course yesterday on linear algebra on the topic "Direct Sum Decompositions of Subspaces and Projections" in the Lecture 32. There I found an equivalence of three statements which are as follows :

Let $W_1,W_2,\cdots,W_k$ be subspaces of a finite dimensional vector space $V$. Then the following statements are equivalent $:$
$(1)$ $W_1,W_2,\cdots,W_k$ are independent.
$(2)$ $\forall j$, $2 \leq j \leq k$ $$W_j \cap (W_1+W_2+\cdots+W_{j-1}) = \{0 \}.$$
$(3)$ If $B_1,B_2,\cdots,B_k$ are bases for $W_1,W_2,\cdots,W_k$ then $B=B_1 \cup B_2 \cup \cdots \cup B_k$ is a basis for $W=W_1+W_2+\cdots+W_k$.

The instructor only proved $(1) \implies (2)$ and left other as an exercise. I have proved $(2) \implies (1)$ which proves the equivalence of $(1)$ and $(2)$ and then I proved $(1) \implies (3)$. But I think $(3) \implies (1)$ cannot be proved. For instance if I take $W=\mathbb R^3$ and take $W_1=\mathrm{span} \{(1,0,0),(0,1,0) \}$,$W_2=\mathrm{span} \{(0,1,0),(0,0,1) \}$ and $W_3=\mathrm{span} \{(0,0,1),(1,0,0) \}$. Then clearly the given condition $(3)$ holds. Though $(1,1,0)+(0,-1,1)+(-1,0,-1)=(0,0,0)$ where $(1,1,0) \in W_1$ ,$(0,-1,1) \in W_2$ and $(-1,0,-1) \in W_3$. This clearly shows that $W_1,W_2,W_3$ are not independent subspaces of $\mathbb R^3$.
So what is wrong in statement $(3)$. Please help me by telling this. It will really help me a lot.
Thank you in advance.

There is an implied quantifier.  The following statement is indeed equivalent to (1) and (2), so I assume it's what was meant:
(3) For every choice of bases $B_1, \ldots, B_k$ of the subspaces $W_1, \ldots, W_k$ the union $B = B_1 \cup \cdots \cup B_k$ is a basis for the sum $W = W_1 + \cdots + W_k$.
Now for (3) $\Rightarrow$ (1) assume $w_1 + \cdots + w_k = 0$ with $w_i \in W_i$.  For each $W_i$ choose a basis $B_i$ such that if $w_i \neq 0$ then $w_i \in B_i$.  The union $B$ of the $B_i$ is a basis for $W$ so, in particular, the nonzero $w_i$ (which are all contained in this union) must be independent.  If any of the $w_i$ were nonzero the expression $w_1 + \cdots + w_k = 0$ would contradict this, so it must be that $w_i = 0$ for all $i$.
