Take a linear system:
$$\dot{x}=Ax+Bu$$
and assume, either that there is a constant affine force $d$ affecting your system (in your example, the valve left open which drains your tank)
$$\dot{x}=Ax+Bu+d$$
or that you want to stabilize your system outside of the origin at $x=r$. In the second case you can introduce the change of variable $z=x-r$ and obtain
$$\dot{z}=Az+Bu+Ar$$
taking $Ar=d$ we notice that the two situations are similar. Both situation will generate a steady state error for a proportional controller.
For the rest of the answer I will demonstrate without loss of generality the steady state error in the first case. Say that the pair $(A,B)$ is stabilizable. It means there exists a matrix $K$ such that $u=Kx$ makes the system exponentially stable at the origin when $d=0$ by imposing that $(A+BK)$ is Hurwitz. Indeed the state trajectory is then given by:
$$x(t) = e^{t(A+BK)}x_0$$
which provides $$\lim_{t\to +\infty}x(t) = 0$$
However, when $d\neq 0$, the state trajectory becomes:
$$x(t) = e^{t(A+BK)}x_0+ \int_{0}^{t}e^{(t-s)(A+BK)}d \,\mbox{d}s \\
= e^{t(A+BK)}x_0+ \int_{0}^{t}e^{s(A+BK)}\mbox{d}s \, d\\
= e^{t(A+BK)}x_0+ (e^{t(A+BK)}-I_n)(A+BK)^{-1}d $$
finally providing
$$\lim_{t\to +\infty}x(t) =- (A+BK)^{-1}d$$
Of course $(A+BK)$ being Hurwitz, $\ker [(A+BK)^{-1}] = \{0\}$, and since $d\neq 0$, this demonstrates the existence of a steady-state error.
Note that in some cases, taking $u=Kx+f$ with a well chosen $f$ will compensate this steady state error and avoid the use of a proportional integral (PI) controller. This solution is however less robust than using a PI controller.