Proving $V$ is isomorphic to $W$ iff $\dim V=\dim W$

Your proof of $\Rightarrow$ is technically correct, but if i was your teacher, I would demand that you explain more clearly why $\{w_1,\dots,w_n\}$. I think you got the idea, but just spend one more sentence on it.

In the other direction, you know that $\dim V=\dim W$. To prove that $V$ and $W$ are isomorphic, construct two bases, one for the space $V$ and one for the space $W$. Then build a linear mapping using the two bases and prove that the linear mapping is bijective.

The left to right implication is basically correct, but I would do it differently, as a consequence of two more general facts.

If $f\colon V\to W$ is a linear map and $\{v_1,\dots,v_n\}$ is a spanning set for $V$, then $\{f(v_1),\dots,f(v_n)\}$ is a spanning set for the image (or range) of $f$. Just observe that, if $v\in V$, then $v=\alpha_1v_1+\dots+\alpha_nv_n$, so $$ f(v)=\alpha_1 f(v_1)+\dots+\alpha_n f(v_n). $$

If $f\colon V\to W$ is an injective linear map and $\{v_1,\dots,v_n\}$ is a linearly independent set, then $\{f(v_1),\dots,f(v_n)\}$ is a linearly independent subset of $W$; indeed, if $$ 0=\alpha_1 f(v_1)+\dots+\alpha_n f(v_n) $$ you can write $$ 0=f(0)=f(\alpha_1v_1+\dots+\alpha_nv_n) $$ and, by injectivity, $\alpha_1v_1+\dots+\alpha_nv_n=0$, so $\alpha_1=\dots=\alpha_n=0$.

Now, let $f\colon V\to W$ be linear and bijective. If $\{v_1,\dots,v_n\}$ is a basis of $V$, it is both a spanning set of $V$ and linearly independent. Then $\{f(v_1),\dots,f(v_n)\}$ is a spanning set of $\operatorname{im}(f)=W$ ($f$ is surjective) and linearly independent ($f$ is injective).

The converse implication follows from the fact that assigning the images for the vectors in a basis of $V$ uniquely defines a linear map. So, if $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_n\}$ are bases of $V$ and $W$, consider the unique linear map $f\colon V\to W$ such that $$ f(v_k)=w_k,\quad k=1,2,\dots,n. $$ Then $f$ is surjective, because a spanning set for the image of $f$ is a basis of $W$ by construction. The rank-nullity theorem says that the null space (kernel) of $f$ has dimension $0$, so $f$ is injective.

The most important point is that you must first know that dimension is well defined (for finitely generated spaces): it that same space has two finite bases, they must contain the same number of vectors. This requires soom work to prove, but is assumed to be known in the problem statement.

Now an the image under an isomorphism of a basis is a basis; this can be immediately deduced from the properties defining a basis. That's your $\Rightarrow$ direction.

Conversely, if both spaces have bases with the same number of elements, there is a bijection $f$ between those bases. Now any map from a basis of a vector space $V$ to a vector space $W$ can be uniquely extended to a linear map $\phi: V\to W$. Applying that to $f$ and then to the inverse bijection $f^{-1}$ between the bases, one gets linear maps $\phi: V\to W$ and $\psi: W\to V$ that are inverses of each other (both compositions linearly extend the identity map on one of the bases, giving the identity operator on its vector space) so $\phi$ and $\psi$ are (mutually inverse) isomorphisms.