$\Bbb{R}^n$ 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?
 A: 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 $n$.
A: 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!
