I'm not assuming any boundary conditions since you didn't specify any.. but an analogous argument will work if there are.
If you separate variables, you get that $u(x,t)$ is of the form $u(x,t) = \sum_{n=0}^{\infty} A_n \cos({n\pi \over L}x)e^{-{n^2\pi^2k^2 \over L^2}t}+ \sum_{n=1}^{\infty} B_n \sin({n\pi \over L}x)e^{-{n^2\pi^2k^2 \over L^2}t} $ Notice that as $t$ goes to infinity, all terms go to zero except the first term of the cosine series. In fact as $t \rightarrow \infty$, $u(x,t)$ converges uniformly to $A_0$, which is the initial average temperature from $x = 0$ to $x = L$. Unless $u(x,0)$ is constant (in which case $u(x,t)$ is constant and there is nothing to prove), there is going to be some $x_0$ for which $|u(x_0,0)| = A_0 > A$.
You can prove a maximal principle as follows. Let $t_0$ be such that $|u(x,t)| < A_0$ for all $t \geq t_0$. Then by the maximal principle on the box $[0,L] \times [0,t_0]$, $|u(x,t)|$ achieves its maximum somewhere on the boundary of the box. On $[0,L] \times [t_0,\infty)$, $|u(x,t)|$, being less than $A_0$, is less than $|u(x_0,0)|$, which is in turn at most the maximum on the boundary of the box.
So $|u(x,t)|$ achieves its overall maximum on the boundary of the box $[0,L] \times [0,t_0]$. It can't occur on the side $[0,L] \times \{t_0\}$ because $|u(x,t)| < A_0$ there. Hence it achieves its maximum on one of the other three sides, which are all part of the boundary of the original domain $[0,L] \times [0,\infty)$. So we conclude that $|u(x,t)|$ does achieve its supremum over $[0,L] \times [0,\infty)$, and on the boundary of that domain.