$V=W_1\oplus\cdots\oplus W_k$ if and only if $\dim(V)=\sum{\dim(W_i)}$ If $W_1,\dots, W_k$ are subspaces of a finite dimensional vector space $V$ such that $W_1+\cdots+W_k=V$, and I want to show that $V=W_1\oplus\cdots\oplus W_k$ if and only if $\dim(V)=\sum{W_i}$, then will what's displayed below suffice?

$$V=W_1\oplus\cdots\oplus W_k$$
$$\iff$$
$$V=W_1+\cdots+W_k~\text{and}~W_i \cap (W_1 + \ldots + W_{i-1} + W_{i+1} + \ldots + W_k) = \{0\}$$
$$\iff$$
$$\text{The subspaces $W_i$ are independent; that is, no sum $w_1+\cdots+w_k$ with $w_i$ in $W_i$ is zero except the trivial sum.}$$
$$\iff$$
$${\scr{B}}=\{\beta_1,\dots,\beta_k\}~\text{is a basis for $V$, where $\beta_i$ is a basis for $W_i$}$$
$$\iff$$
$$\dim{V}=\dim{(W_1+\cdots+W_k)}=\dim{W_1}+\cdots+\dim{W_k}=~\mid\beta_1\mid+\cdots+\mid\beta_k\mid=k$$
$$\iff$$
$$\overset{\text{Does this belong here?}}{\dim{\scr{B}}=k}$$
 A: Your first equivalence is false: when $k \geq 3$, $\bigcap_{i=1}^k W_k = \{0\}$ is too weak to imply that the spaces $W_1,\ldots,w_k$ are independent, i.e., that $W_1 + \ldots + W_k = W_1 \oplus \ldots \oplus W_k$.  For example, take $V = \mathbb{R}^2$, $W_1 = \langle (1,0) \rangle$, $W_2 = \langle (0,1) \rangle$, $W_3 = \langle (1,1) \rangle$.  This is one of two or three subtle traps in linear algebra that even professional mathematicians can fall into if they're not careful.  
Any of the following is an acceptable definition of independent subspaces (and they are equivalent):  
(i) For all $1 \leq i \leq k$, $W_i \cap (W_1 + \ldots + W_{i-1} + W_{i+1} + \ldots + W_k) = \{0\}$.
(ii) If for all $i$ we choose a nonzero $v_i \in W_i$, then $\{v_1,\ldots,v_k\}$ is a linearly independent set.
(iii) If for all $i$ we choose a linearly independent set $S_i \subset W_i$, then 
$S = \bigcup_{i=1}^k S_i$ is a linearly independent set.
Note that if you believe that (iii) is equivalent to the sum being a direct sum then you've got the equivalence you're asking about, so I suggest you concentrate instead on showing these conditions are equivalent to your given definition of internal direct sums.  
