Prove using superposition that an LTI system is linear
I'm going to assume the system has the initial condition $x(0) = 0$, as otherwise the system isn't linear. 

Suppose the input $u_1(t)$ yields the output $x_1(t)$ and the input $u_2(t)$ yields the output $x_2(t)$. 

We need to prove that the input $u(t) = \alpha u_1(t)+\beta u_2(t)$ yields the output $x(t) = \alpha x_1(t) + \beta x_2(t)$ for any scalars $\alpha$ and $\beta$. 

The given input and output pairs must satisfy the differential equation and the initial condition. Hence, $x_1'(t) = Ax_1(t)+Bu_1(t)$ and $x_2'(t) = Ax_2(t)+Bu_2(t)$ for all times $t$ and $x_1(0) = x_2(0) = 0$.

Now, add $\alpha$ times the first equation to $\beta$ times the second equation to get: $\alpha x_1'(t) + \beta x_2'(t) = \alpha(Ax_1(t)+Bu_1(t)) + \beta(Ax_2(t)+Bu_2(t))$

Then, rearrange stuff to get:

$\dfrac{d}{dt}[\alpha x_1(t) + \beta x_2(t)] = A(\alpha x_1(t)+\beta x_2(t)) + B(\alpha u_1(t)+\beta u_2(t))$  

$x'(t) = Ax(t)+Bu(t)$

Also, we have $x(0) = \alpha x_1(0)+\beta x_2(0) = \alpha \cdot 0 + \beta \cdot 0 = 0$. 

Thus, the signals $u(t) = \alpha u_1(t)+\beta u_2(t)$ and $x(t) = \alpha x_1(t) + \beta x_2(t)$ satisfy the differential equation and the initial condition. Hence, the input $u(t) = \alpha u_1(t)+\beta u_2(t)$ yields the output $x(t) = \alpha x_1(t) + \beta x_2(t)$ for any inputs $u_1(t)$ and $u_2(t)$ and any scalars $\alpha$ and $\beta$. 

Therefore, the system is linear, as desired.