Prove that any linear functional on $\mathbb{R^{n}}$ with the Euclidean norm is bounded I'm not sure how to prove this without just using the fact that is equivalent to say that a linear functional is continous.
 A: For
$\vec v \in \Bbb R^n, \tag 1$
we may write
$\vec v = \displaystyle \sum_1^n v_i \vec e_i, \tag 2$
where $\vec e_i$, $1 \le i \le n$ is a basis of $\Bbb R^n$, orthonormal in the Euclidean inner product $\langle \cdot, \cdot \rangle_2$ on $\Bbb R^n$, which defines the Euclidean norm norm $\Vert \cdot \Vert_2 = \langle \cdot, \cdot \rangle^{1/2}$ on $\Bbb R^n$ (see note below); then with 
$\phi: \Bbb R^n \to \Bbb R \tag 3$
linear, we have
$\phi(\vec v) = \phi \left (\displaystyle \sum_1^n v_i \vec e_i \right ) = \displaystyle \sum_1^n v_i \phi(\vec e_i); \tag 4$
thus
$\vert \phi(\vec v) \vert = \left \vert \displaystyle \sum_1^n v_i \phi(\vec e_i) \right \vert \le \displaystyle \sum_1^n \vert v_i \phi(\vec e_i) \vert =  \sum_1^n \vert v_i \vert \vert \phi(\vec e_i) \vert; \tag 5$
let
$M = \max \{\vert \phi(\vec e_i) \vert, \; 1 \le i \le n \}; \tag 6$
then
$\vert \phi(\vec v) \vert \le \displaystyle \sum_1^n \vert v_i \vert \vert \phi(\vec e_i) \vert \le M \sum_1^n \vert v_i \vert; \tag 7$
also,
$\vert v_i \vert \le \sqrt { \displaystyle \sum_1^n v_i^2 }, \ 1 \le i \le n; \tag 8$
whence
$\displaystyle \sum_1^n \vert v_i \vert \le n \sqrt { \displaystyle \sum_1^n v_i^2 }, \tag 9$
so that (7) becomes
$\vert \phi(\vec v) \vert \le M n \sqrt { \displaystyle \sum_1^n v_i^2 } = Mn\Vert \vec v \Vert_2, \tag {10}$
which shows that $\phi$ is bounded in the Euclidean norm $\Vert \cdot \Vert_2$ on $\Bbb R^n$, with bound $Mn$.
Nota Bene:  The definition of Euclidean norm $\Vert \cdot \Vert_2$ on $\Bbb R^n$ used here is the usual one defined in terms of the Euclidean inner product $\langle \cdot, \cdot \rangle_2$, where if
$\vec w = \displaystyle \sum_1^n w_i \vec e_i, \tag{11}$
we have
$\langle \vec v, \vec w \rangle_2 = \displaystyle \sum_1^n v_i w_i, \tag{12}$
from which
$\Vert \vec v \Vert_2 = \sqrt{\langle \vec v, \vec v \rangle_2} = \sqrt {\displaystyle \sum_1^n v_i^2}; \tag{13}$
the vectors $\vec e_i$ are chosen orthonormal with respect to this basis:
$\langle \vec e_i, \vec e_j \rangle = \delta_{ij}, \; 1 \le i, j \le n. \tag{14}$
End of Note.
