$\{\mathbf 0
\}\subseteq W_j\cap W_l\subseteq W_j\cap\sum_{i\not=j}W_i=\{\mathbf 0\}$ for all $l\not=j$. Thus also $\bigcap_{i=1}^k W_i=\{\mathbf 0\}$.
Moreover, let $V=\mathbb{R}^2$, $W_1=\mathbb{R}(1,0)$, $W_2=\mathbb{R}(0,1)$, $W_3=\mathbb{R}(1,1)$. The $W_1\cap W_2\cap W_3=\{(0,0)\}$, but $W_3\cap (W_1+W_2)=W_3\not=\{(0,0)\}$.
The fact that $W_1+W_2=W_2+W_1$ is fairly obvious, because
$$
W_1+W_2=L(W_1\cup W_2)=L(W_2\cup W_1)=W_2+W_1
$$
by the very definition.
What about associativity? In this case you use the proposition: if $W_1,W_2,W_3$ are subspaces, $X=W_1+W_2$ and $Y=W_2+W_3$, you want to prove that
$$
X+W_3=W_1+Y
$$
Let $x\in X,w_3\in W_3$; then, by the proposition, $x=w_1+w_2$, with $w_1\in W_1$, $w_2\in W_2$; then
$$
x+w_3=(w_1+w_2)+w_3=w_1+(w_2+w_3)\in W_1+Y
$$
because $w_2+w_3\in Y$. Thus $X+W_3\subseteq W_1+Y$. The reverse inclusion follows similarly.
About direct sums there is a big misunderstanding. While the definition of “direct sum” in the case of two subspaces is correct, it is incorrect to say that the sum of more than two subspaces is direct when $W_i\cap W_j=\{0\}$ for $i\ne j$.
The condition is stricter, namely that
$$
W_i\cap\sum_{j\ne i}W_j=\{0\},\qquad i=1,2,\dots,n
$$
at least if one wants to stick with the common terminology and one of the most important properties of direct sums, namely that
$$
\dim(W_1\oplus W_2\oplus\dots\oplus W_n)=\dim W_1+\dim W_2+\dots+\dim W_n
$$
in case of finite dimensional spaces.
For instance, the enclosing vector space being $\mathbb{R}^3$, if $W_1$ is generated by $(1,0,0)$, $W_2$ by $(0,1,0)$ and $W_3$ by $(1,1,0)$, it is true that $W_1\cap W_2=\{0\}$, $W_1\cap W_3=\{0\}$, $W_2\cap W_3=\{0\}$, but
$$
\dim(W_1+W_2+W_3)=2\ne\dim W_1+\dim W_2+\dim W_3
$$
In any case, since a direct sum is a sum of subspaces to begin with, proving associativity (once the definition is fixed) and commutativity is not a problem, because it has already been done.
Let's tackle associativity of direct sum. Suppose $W_1,W_2,W_3$ are independent subspaces (meaning that their sum is direct). Then, by definition,
$$
W_1\cap(W_2+W_3)=\{0\}=(W_1+W_2)\cap W_3
$$
Therefore also $W_1\cap W_2=\{0\}=W_2\cap W_3$; hence $W_1+W_2=W_1\oplus W_2$ and $W_2+W_3=W_2\oplus W_3$. Hence
$$
W_1+(W_2+W_3)=W_1\oplus(W_2+W_3)=W_1\oplus(W_2\oplus W_3)
$$
$$
(W_1+W_2)+W_3=(W_1\oplus W_2)+W_3=(W_1\oplus W_2)\oplus W_3
$$
But these are equal by the previous argument.
Best Answer
There are several points you need to consider.
First, what do you mean by invariant subspaces? Invariance only has meaning with respect to some transformation. But this should have no bearing on the question.
Second, what do you mean by isomorphism? If you mean as linear spaces, then it is enough to show a bijective linear function between $\frac{W_1 \oplus W_2 \oplus ... \oplus W_n}{W_i}$ and $W_1 \oplus W_2 \oplus ... W_{i-1}\oplus W_{i+1} \oplus W_n$.
Keeping this in mind, it is enough to show that if $V = W \oplus U$ then $V/W \simeq U$ (treat $U$ as $W_1 \oplus W_2 \oplus ... W_{i-1}\oplus W_{i+1} \oplus W_n$ and $W$ as $W_i$).
In this case, if $f$ is the linear projection $f(w,u) = (0,u)$ then since $ker(f) = W$, $f$ may be considered as a function from $V/W$ by $\tilde{f}(u + W) = u$. Clearly $\tilde{f}$ is onto since for any $u \in U$, $\tilde{f}(u + W) = u$. Also, $\tilde{f}$ is bijective since for any $u \neq 0$, $\tilde{f}(u + W) = u \neq 0$. Hence, $\tilde{f}$ is a bijective linear map between $V/W$ and $U$ as required.