Some problematic points in your proof:
Proof: let $\{a_1,a_2,...,a_i,v_1,v_2...v_n\}$ be basis for $W_1$, then $\dim(W_1)= i+n$ and $\{b_1,b_2,...,b_j,v_1,v_2...v_n\}$ be basis for $W_2$, then $\dim(W_2)= j+n$ .
You are missing some fundamental information. What are $i,j,n$? Why do both bases contain the same vectors $v_1,\dots,v_n$?
the definition of sum of two subspaces tells us that the basis of the sum is a combination of those two subspaces,
Presumably, you mean "a combination of those two bases". In any case, the term "a combination of" is too vague for this statement to be correct.
which is $\dim(W_1+W_2)= i +j+n$. Hence we can arrive that $W_1+W_2$ is finite-dimensional.
Since the both subspaces have n elements in common, so $\dim(W_1 \cap W_2)= n$.
It is not true that the two subspace have $n$ elements in common. If we're talking about vector spaces over $\Bbb R$ or $\Bbb C$, then the subspaces should have either infinitely many elements or one element in common.
A correct proof, in which I have attempted to parallel yours as much as possible.
Let $v_1,\dots,v_n$ be a basis of $W_1 \cap W_2$. Since $W_1 \cap W_2 \subseteq W_1$, we can extend this to a basis $v_1,\dots,v_n,a_1,\dots,a_i$ of $W_1$. Similarly, let $v_1,\dots,v_n,b_1,\dots,b_j$ be a basis of $W_2$. It is clear that the union of these bases,
$$
\mathcal B = \{v_1,\dots,v_n,a_1,\dots,a_i,b_1,\dots,b_j\}
$$
is a spanning set of $W_1 + W_2$. In order to show that this is a basis, we must also show that $\mathcal B$ is linearly independent.
One we have proven the claim that $\mathcal B$ is indeed a basis, we may simply count the elements of each basis to find
$$
\dim(W_1 \cap W_2) = n, \quad \dim(W_1) = n+i, \quad \\
\dim(W_2) = n+j, \quad \dim(W_1 + W_2) = n+i+j.
$$
We can then verify the desired result by plugging these in to the desired equation.
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
Your proof is not right. $W_1\cap(W_2+W_3)$ is not generally equal to $W_1\cap W_2+W_1\cap W_3$. Consider $W_1, W_2, W_3$ - three different subspaces of dimension $1$ in a vector space $V$ of dimension $2$. (Imagine three different lines through the origin in a plane). There you have $W_1\cap W_2=W_1\cap W_3=\{0\}$ but $W_1\cap(W_2+W_3)=W_1\cap V=W_1$.
Of course, the equality is true when $W_2\subset W_1$ - but this is pretty much equivalent to your question, so your argument would be circular (i.e. you would be assuming the very thing you are asked here to prove).
Note: I wrote this answer before you changed the question from "check if my proof is right" to "check if my proof is right, and if not, show me the right proof". It the meantime there have been useful hints in other answers, which I won't repeat. The basic idea is - $W_1\cap(W_2+W_3)=W_2+(W_1\cap W_3)$ is an equality of sets, so you can approach it as any other set equality: suppose $x$ is in the LHS, prove that it belongs to RHS and vice versa. One direction should be a bit easier than the other. Give it a try.