One shows that $v_1,v_2,v_3$ is a linearly independent set of vectors. Since there are $3$ of them and $\Bbb R^3$ is a $3$-dimensional vector space they must constitute a basis of $\Bbb R^3$.

Suppose they are not linearly independent, meaning there is are some constants $\mu_1,\mu_2,\mu_3\in\Bbb R$ so that:

$$\mu_1 v_1+\mu_2v_2+\mu_3v_3=0$$and that not all of the $\mu_i$ are $0$.

You then get

$$\mu_1 v_1 = -\mu_2 v_2 - \mu_3 v_3$$

applying $A$ to this gives:

$$A(\mu_1v_1)=\lambda_1 \mu_1v_1 = -\mu_2A(v_2)-\mu_3 A(v_3) = -\mu_2\lambda_2 v_2 -\mu_3\lambda_3v_3$$

but you also have that

$$\lambda_1 \mu_1 v_1 = \lambda_1 (-\mu_2 v_2- \mu_3v_3)=-\lambda_1\mu_2 v_2 - \lambda_1\mu_3v_3$$

whence:

$$0=\lambda_1\mu_1 v_1 - \lambda_1\mu_1v_1 = - (\lambda_1-\lambda_2)\mu_2 v_2 -(\lambda_1-\lambda_3)\mu_3 v_3$$

in other words $\xi_2 v_2 = \xi_3 v_3$ for $\xi_2= (\lambda_1-\lambda_2)\mu_2$ and $\xi_3= -(\lambda_1-\lambda_3)\mu_3$. Apply $A$ again to both side to get:

$$A(\xi_2v_2) = \lambda_2 \xi_2 v_2 \overset!= A(\xi_3 v_3) = \lambda_3( \xi_3v_3) = \lambda_3 \xi_2 v_2$$

whence:

$$(\lambda_2-\lambda_3)\xi_2v_2=0$$

since $\lambda_2\neq\lambda_3$ you get that $\xi_2=0$, it follows that from $\xi_2 v_2 = \xi_3v_3$ that $\xi_3=0$ also. Now remember that:

$$\xi_i = (\lambda_1-\lambda_i)\mu_i$$

since $\lambda_1-\lambda_i\neq0$ for $i\in\{1,2\}$ you get that $\mu_i=0$ for $i\in\{1,2\}$. Then starting with our first equation we get:

$$\mu_1 v_1 +\underbrace{\mu_2v_2+\mu_3v_3}_{=0}=0,\implies \mu_1v_1=0$$

and $\mu_1=0$ also. This contradicts our assumption that not all of the $\mu_i$ are $0$, meaning that our assumption that $v_1,v_2,v_3$ are not linearly independent is false.