Simple proof for uniqueness of solutions of linear ODEs? Consider the system of linear ODEs $\dot{x}(t)=Ax(t)$, $x(0)=x_0\in\mathbb{R}^n$. Does anyone know a simple proof showing that the solutions are unique that does not require resorting to more general existence/uniqueness results (e.g., those relating to the Picard iteration) nor solving for the solutions explicitly?

Since the students are engineers, why don't you want to show them explicit solutions, which surely they'd need to see anyway?  If we knew about a matrix exponential $e^{At}$, then to show $x(t) = e^{At}x_0$ let's look at the $t$-derivative of $e^{-At}x(t)$, which is 
$$
e^{-At}x'(t) + (-Ae^{-At})x(t) = e^{-At}Ax(t) - Ae^{-At}x(t).
$$
From the series definition of the matrix exponential, $A$ and $e^{Bt}$ commute if $A$ and $B$ commute, so $A$ and $e^{-At}$ commute. Thus
$$
(e^{-At}x(t))' = e^{-At}Ax(t) - Ae^{-At}x(t) = Ae^{-At}x(t) - Ae^{-At}x(t) = 0.
$$
Therefore $e^{-At}x(t)$ is a constant vector, and setting $t = 0$ tells us this constant vector has to be $x(0) = x_0$. Thus $e^{-At}x(t) = x_0$, so $x(t) = e^{At}x_0$ if we know that $e^{At}$ and $e^{-At}$ are inverses of each other.
Note that this solution can be thought of as a higher-dimensional version of the integration-free proof that the only solution of the 1-dim. ODE $y'(t) = ay(t)$ with $y(0) = y_0$ is $y_0e^{at}$: if $y(t)$ is a solution then the derivative of $e^{-at}y(t)$ is
$$
e^{-at}y'(t) - ae^{-at}y(t) = e^{-at}(ay(t)) - ae^{-at}y(t) = 0.
$$
Thus $e^{-at}y(t)$ is a constant function, and at $t = 0$ we see the value is $y(0) = y_0$, so $e^{-at}y(t) = y_0$. Thus $y(t) = y_0e^{at}$. In higher dimensions we just need to be more careful about the order of multiplication (e.g., the way the product rule is formulated for matrix-valued functions). 
