If $V=W_1+\cdots+W_k$ and $\dim(V) =\sum_{i=1}^{k}\dim(W_i)$, prove that $V = W_1 \oplus\cdots\oplus W_k$ 
Let $V$ be a vector space and let $W_1,\ldots,W_k$ be subspaces of $V$ such that $V = W_1 +\cdots+ W_k$ and $\dim(V) = \dim(W_1) +\cdots+ \dim(W_k)$. Prove that $V = W_1 \oplus\cdots\oplus W_k $.

My attempt: 
$k=1$ is trivial, for $k=2$, 
$\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ which gives us $\dim(W_1 \cap W_2)=0$, which implies $W_1 \cap W_2=\{0\}$. Hence $V = W_1 \oplus W_2 $. 
For $k=3$, we get $\dim(W_1+W_2+W_3)=\dim(W_1+W_2)+\dim(W_3)-\dim((W_1+ W_2)\cap W_3)$, which implies $\dim(W_1 \cap W_2)+\dim((W_1+ W_2)\cap W_3)=0$. We get the following: 
$1. \,W_1 \cap W_2 = \{ 0 \}$
$2. \, (W_1+ W_2)\cap W_3 = \{0\} $ , $(W_1+ W_3)\cap W_2 = \{0\} $ and $(W_2+ W_3)\cap W_1 = \{0\}$
Are conditions $1$ and $2$ enough to prove that $V = W_1 \oplus W_2 \oplus W_3  $ ? This method isn't good for solving this problem for general $k$, is there any better way to prove this?

You have a map $T$ from the external direct sum $W_1\oplus\cdots W_k$
to $V$ taking $(w_1,\ldots,w_k)$ to $w_1+\cdots+w_k$. Its image
is $W_1+\cdots+W_k=V$. But both $W_1\oplus\cdots W_k$ and $V$ have the same
dimension, so $T$ is an isomorphism. The injectivity of $T$ implies
that inside $V$, $W_1+\cdots+W_k$ is an internal direct sum.
