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?
Best Answer
The answer is yes. But you don't need they have the same basis.
More precisely:
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.