Since this is such basic stuff, you must really be clear about what definitions you are using. In the courses I know (and give) "finite dimensional" is actually defined before dimension is, and it means "finitely generated", in other words a space is finite dimensional if (and only if) it is the span of a finite set. Span of a set of vectors is formed by the set of their linear combinations, which forms the smallest subspace containing that set. From this it is easy to show that the span of $A\cup B$ is the sum of the span of $A$ and the span of $B$ (which sum can be defined as the span of their union). Once this is done, it is clear that the sum of two finite dimensional subspaces (spans of finite $A,B$ respectively) is finite dimensional (the span of the finite set $A\cup B$).
On the other hand it is not so obvious (though true) that any subspace of a finite dimensional space is finite dimensional, and in particular that the intersection of a finite dimensional subspace with another subspace (which may or may not be finite dimensional) is finite dimensional.
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.
Best Answer
We know that if U and V are two subspaces of a finite-dimensional vector space: $\dim (U+V)=\dim U + \dim V -\dim (U\cap V)$. To prove the equality requested if suffice to show that $\dim (W_1^0+W_2^0)=\dim (W_1\cap W_2)^0$. This is a direct consequent from the previous equality and the fact that $W_1^0\cap W_2^0=(W_1+ W_2)^0$ and $\dim U^0=\dim V-\dim U$ for all subspace $U$ of $V$.