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

My attempt:

Let $c=$ cardinality of $\Bbb{R}$. Then we know that $\Bbb{R}$ over $\Bbb{Q}$ has basis of cardinality $c$. And for a basis $\mathcal{B}$ of $\Bbb{R}^n$ over $\Bbb{Q}$, we have has $\aleph_0<|\mathcal{B}|\leq |\Bbb{R}^n|=c$. So from Cantor's  hypothesis $|\mathcal{B}|=c$. Therefore bases for $\Bbb{R}^n$ and $\Bbb{R}$ have same cardinality when it is considered as a vector space over $\Bbb{Q}$. Thus they are isomorphic.

Is this proof logically flawless? Please correct me if required. Thank you.
Edit I see use of Cantor's hypothesis is not much justified. So how can we prove that the bases of both vector space have same cardinality without using this hypothesis?

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 $B \to  B\times\{0\} \cup \{0\} \times B$. 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 $n$.
