Let $\{b_1, \ldots, b_k\}$ be a basis for $U \cap W$. Then first you need to argue you can extend this basis to a basis $b_1, \ldots, b_k, c_1, \ldots, c_l$ of $U$ and $b_1, \ldots, b_k, d_1, \ldots, d_n$ of $W$.
Now argue that $b_1, \ldots, b_k, c_1, \ldots, c_l, d_1, \ldots, d_n$ is a basis of $U+W$, so $\dim(U) + \dim(W) = (k + l) + (k + n) = (k + l + n) + k = \dim(U+W) + \dim(U \cap W)$.
I'll gladly fill in any details you are unclear about.
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
Your proof is correct. You can also use proof by contradiction.
Suppose, for a contradiction, that $U\cap W=\{0\}$. Then $\dim(U\cap W)=0$. Since $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$, we have $\dim(U+W)=\dim U+\dim W$. We know that $\dim V\geq\dim(U+W)$, because $U+W$ is a subspace of $V$. It follows that $\dim V\geq\dim U+\dim W$, which contradicts our assumption that $\dim V<\dim U+\dim W$. This completes the proof.