$Bbb{R}^n$ and $Bbb{R}$ are isomorphic as vector spaces over $Bbb{Q}$.

Your proof is correct if you accept the axiom of choice which assures the existence of a basis $mathcal B_k$ of $mathbb R^k$ as a vector space over $mathbb Q$. Then $mathcal B_n$ and $mathcal B_1$ have the same cardinality. Now you invoke the following lemma:

Given vector spaces (over any field) $V_i$ with bases $mathcal B_i$ having the same cardinality. Then $V_1$ and $V_2$ are isomorphic.

Now cearly each function $f : mathcal B_1 to V_2$ extends to a unique linear map $f' : V_1 to V_2$. Now take a bijection $phi : mathcal B_1 to mathcal B_2$ and define $f(x) = phi(x)$. Then it is easy to verify that $f'$ is a linear isomorphism. Its inverse is induced by $phi^{-1}$.

Edited:

Let $B$ a basis for $V$. Then it is easy to show that the set of all pairs having the form $(b,0)$ and $(0,b')$ with $b,b' in B$ form a basis for $V times V$.

It is a well-known consequence of the axiom of choice that if $B$ is infinite, then there exists a bijection $B to Btimes{0} cup {0} times B$. Note that if $B$ is countable, then induction is sufficient to show this.

This shows that if $mathcal B$ is infinite, then $V$ and $V times V$ are isomorphic. Hence also $V$ and $V^n = V times ldots times V$ are isomorphic for any $n$.

Note that $c$ depends on the underlying field ($mathbb{Q}$). The existence of a basis requires the axiom of choice. I do not know if the statement is correct. You are using the same cardinality to establish isomorphism property. This does only hold for finite-dimensional vectir spaces!