Really a comment, unfortunately too long for the comment box:

Your terminology is incorrect: the theory of distributions always requires the "domains" to be open. So here your $\Omega$ is an open set, fixed once and for all as the "space domain" of your given distribution $u\in W^1$, and it really makes no sense to consider "$\Omega$ compact" when you speak of $u$. This being said, you still have to decide whether your function $f$ is in $\mathcal{C}^1([0,T]\times\Omega)$ or in the better space $\mathcal{C}^1([0,T]\times\overline{\Omega})$. At this point open/compact makes both sense and a real difference when speaking of $f$ because you assume it is $\mathcal{C}^1$ in the classical sense, so I guess this is what you meant by "$\Omega$ compact".

In the easy case $f\in\mathcal{C}^1([0,T]\times\overline{\Omega})$ the answer to your question is yes as you pointed out, basically because $f\in\mathcal{C}^1([0,T]\times\overline{\Omega})$ is uniformly bounded together with all its derivatives. If you have no boundary regularity in space, the answer is no in general: take for example $u\equiv 1$ in $[0,T]\times \Omega$ (which is indeed a nice element of your $W^1$ space!) and choose any $f=f(x)$ to be constant in time, belonging to $\mathcal{C}^1(\Omega)$ but not to $\mathcal{C}^1(\overline{\Omega})$, and such that $f\notin L^2(\Omega)$ (e.g. blow up at the boundary). Clearly $u,f$ satisfy your hypotheses, but the product $fu$ is not even $L^2$ in space because $f$ is not.

Have you tried googling something like "product rule in Sobolev spaces"?