The error lies in assuming that the only possibility for the supplementary subspace is $\langle v_{m+1},\ldots,v_n\rangle$. It isn't. If $V=\mathbb{R}^2$, $U=\langle(1,0)\rangle$ and if you complete the basis, which vector will you take? You will choose $(0,1)$ probably. And indeed $\langle(0,1)\rangle$ is a supplementary subspace. However, $\langle(1,1)\rangle(=\langle(1,0)+(0,1)\rangle)$ will also work, for instance.
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
In your attempt, while the sum $W_1 + W_2$ is direct (that is, $W_1 \cap W_2 = 0$), it might not equal all of $W$.
There is a simple counterexample for $V = \mathbb{R}^2$. $V$ is the direct sum of the $x$ and $y$ axes, but if we set $W$ equal to the line $y = x$, then $W$ intersected with the $x$ and $y$-axes is zero.