Why is the state space representation, $dx/dt = A x(t) + B u(t)$ linear?
What you have listed is a state equation.  Usually a state-space representation has two equations:

(i) $\dot{x}(t)=Ax(t)+Bu(t)$
(ii) $y(t)=Cx(t)+Du(t)$

One takes the input $u(t)$ and generates the state $x(t)$ and another that takes the input $u(t)$ and state $x(t)$ and generates the output $y(t)$.  You are missing the second equation.

That aside, the equation that you have is basically a 1st order linear differential equation.  It's important to note, here, that the input is not $x(t)$, it is $u(t)$.  $x(t)$ is the state variable.

I will examine just the first half (input to state) and leave the second half (input and state to output) to you.  I will remark, though, that if $x(t)$ is linear with respect to $u(t)$ then $y$ in equation (ii) above necessarily has to be linear by virtual of adding two linear terms together, so the second half is really trivial.

The first half is:

(1) $\dot{x}(t)=Ax(t)+Bu(t)$

Can you show that it is linear?  

I will show additivity and leave homogeneity to you.  Say you have inputs $u_1$, $u_2$, and $u_3=u_1+u_2$.

One way to do this is to solve (1).  As it is a 1st order ODE, it is not too difficult, but let's be lazy.

Let $x_1$ be the state generated by input $u_1$ and $x_2$ be the state generated from input $u_2$, then we have

(2) $\dot{x_1}(t)=Ax_1(t)+Bu_1(t)$

and

(3) $\dot{x_2}(t)=Ax_2(t)+Bu_2(t)$

adding (2) and (3) gives

(4) $\dot{x_1}(t)+\dot{x_2}(t)=Ax_1(t)+Ax_2(t)+Bu_1(t)+Bu_2(t)$

From linearity of derivative, we get that:

(5) $\dot{x_3}(t)=A(x_3(t))+B(u_3(t))$

where $x_3 = x_1+x_2$.

So input $u_3 = u_1+u_2$ generates state variable $x_3 = x_1+x_2$, showing that additivity holds for the state equation.