The matrix $A$ takes a vector in the subspace $W$, considered as a copy of $\mathbb R^k$, expressed in the basis $B$, and returns the corresponding vector expressed in the basis $a$, as a member of $\mathbb R^n$.
The transpose of $A$ does the opposite: it takes a vector in $\mathbb R^n$, expressed in the basis $a$, and gives a vector that is intrinsic to the subspace $W$. This new vector is expressed in the basis $B$. This new vector, necessarily then, has lost any information about components outside of $W$.
So the transpose takes a general vector and projects it onto $W$, but you then only have that new vector expressed in terms of the the basis $B$. The original matrix $A$ converts any such vector back to the $a$ basis.
What you should show in order to prove this result is that any out-of-subspace components of an arbitrary input vector are reduced to zero as a result of this composition. You will need the images of the vectors $w_1, w_2, \ldots$ under $A^T$ to be able to decompose any arbitrary vector of $\mathbb R^n$ this way.
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
If you know matrices, this will do it: Regard $P$, $v$, and $w$ are column vectors. Let $M$ be the matrix whose two columns are $v$ and $w$. It's an $n\times2$ matrix. (By the way, you shouldn't use capital $N$ and lower-case $n$ as if they were synonymous. Mathematical notation is case-sensitive.) Then $M^TM$ is a $2\times 2$ matrix, which is invertible if the vectors $v$, $w$ are linearly independent. The matrix $M(M^TM)^{-1}M^T$ is and $n\times n$ matrix of rank $2$. The vector $M(M^TM)^{-1}M^T P$ is the projection that you seek.
"Usage note": Once upon a time a highly respected and moderately famous mathematician told me that $M(M^TM)^{-1}M^T$ is the identity matrix. Apparently he was assuming $M$ was a square matrix. I have a bold hypothesis, which I haven't checked empirically: "Pure" mathematicians tacitly assume matrices are square; "applied" mathematicians don't.