General solution from distinct real eigenvalues

First check $e^{\lambda_i t}\mathbf{v}_i$ solves the system: its derivative is $\lambda_i e^{\lambda_i t}\mathbf{v}_i$, while $A(e^{\lambda_i t}\mathbf{v}_i)=e^{\lambda_i t}A\mathbf{v}_i=e^{\lambda_i t}\lambda_i\mathbf{v}_i$ using $A\mathbf{v}=\lambda\mathbf{v}$; the two sides match, so it is indeed a solution for $i=1,2$.

Linear independence of $\mathbf{v}_1,\mathbf{v}_2$: suppose $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{0}$. Applying $A$ gives $c_1\lambda_1\mathbf{v}_1+c_2\lambda_2\mathbf{v}_2=\mathbf{0}$. Multiplying the first relation by $\lambda_2$ and subtracting gives $c_1(\lambda_1-\lambda_2)\mathbf{v}_1=\mathbf{0}$; since $\lambda_1\neq\lambda_2$ and $\mathbf{v}_1\neq\mathbf{0}$, we get $c_1=0$, and then $c_2=0$ too. So $\mathbf{v}_1,\mathbf{v}_2$ are independent.

By linearity of the system (exactly as in the scalar case), any combination $c_1e^{\lambda_1t}\mathbf{v}_1+c_2e^{\lambda_2t}\mathbf{v}_2$ solves the system. Because $\mathbf{v}_1,\mathbf{v}_2$ are independent, matching any initial condition $\mathbf{x}(0)=\mathbf{x}_0$ means solving $c_1\mathbf{v}_1+c_2\mathbf{v}_2=\mathbf{x}_0$ for $(c_1,c_2)$, which has a unique solution since $\{\mathbf{v}_1,\mathbf{v}_2\}$ is a basis of the plane. So every solution is captured, giving the general solution $\mathbf{x}(t)=c_1e^{\lambda_1t}\mathbf{v}_1+c_2e^{\lambda_2t}\mathbf{v}_2$.