Two vector spaces with same dimension and same basis, are identical? Let $V$ subspace of $W$ and both have same dimension and same basis. Then can we safely say that $V= W$ ? 
I believe yes. For example there may be an element $x \in V$ written as a linear combination of the basis elements. This linear combination is unique. 
Now let's take another vector $y \in W$. The $y$ is written as a linear combination of the basis vectors. 
We equate the two linear combinations, and since the basis elements are linearly independent, we get that $x=y$. 
Thus $V=W$. 
Do you agree?
 A: The answer is yes. But you don't need they have the same basis.
More precisely:

Let $V$ be a subspace of the finite dimensional space $W$; if $\dim V=\dim W$, then $V=W$.

Proof. Since $\dim V\le \dim W$ for any subspace $V$ of $W$ we can prove the equivalent statement that if $V\subsetneq W$ (proper subspace), then $\dim V<\dim W$. If $\{v_1,v_2,\dots,v_n\}$ is a basis for $V$ and $w\in W$ but $w\notin V$, it is easy to show that $\{v_1,v_2,\dots,v_n,w\}$ is linearly independent (prove it). However, any linearly independent subset of $W$ can be extended to a basis, implying $\dim W\ge n+1$.

Your attempt shows you have the right idea, but express it poorly.
Let $w\in W$. Since $w$ is a linear combination of the common basis of $V$ and $W$, we conclude that $w\in V$. Together with the assumption that $V\subseteq W$, this ends the proof.
A: Here is another proof. We know that $V$ is a subspace of $W$ and $\dim(V) = \dim(W)$. Now suppose that V is a proper subset of W. We know that V must have a complement that is a subspace of $W$. Let that complement be $U$. We have that:
$$U + V = W\\U\cap V = \varnothing$$ 
By Greenman's theorem, we have: 
$$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$ which, using the relations above simplifies to:
$$\dim(W) = \dim(U) + \dim(V) $$
Now, it is obvious that $\dim(U) \ge 1$: Suppose $\dim(V) = 0$, then $V = \varnothing$. But this is impossible since $U + \varnothing = U $, which contradicts the fact that $U$ is a complement of $V$ in $W$. 
It follows immediately that $\dim(V) < \dim(W)$. This is a contradiction. Therefore $V$ is not a proper subset of $W$, in other words:
$$V = W$$
A: Note that saying $V,W$ as having the same basis and same dimension is somewhat redundant; you should think and phrase that sentence as '$V,W$ are both generated by a basis of the same dimension', having the same basis for instance will directly imply they have the same dimension.
If they have the same basis then clearly they are the same, since every element is generated by it.
So, if indeed $V \leq W$  and both are generated by a basis of the same dimension then $V = span\{v_{1},..,v_{n}\}$ for $v_{i} \in W$, and indeed $\exists \{v_{1},..,v_{n},v_{n+1},..v_{m}\} \subset W$ completed to a basis of $W$, assuming the dimensions are finite.
But since we assume $dim V = dim W$ then $m = n$ and any addition to $\{v_{1},..,v_{n}\}$ results in a linearly dependent set within $W$. Then $span\{v_{1},..,v_{n}\} = W = V$.
If $dim V = dim W \approx N$, for example, and $V \leq W$ take:
$V = span\{cos(nx) | n \in N\} < span\{sin(nx),cos(nx) | n \in N \} = W$ to see how this proposition doesn't hold
