is $F(x, n)=x(n)u(n)$ time invariant?
The answer to this (and your other similar question) is most likely found in the somewhat unusual notation used for the system.

In ordinary terms, if you just take your system as a map between input and output signal, then you're absolutely right, it would not be time invariant.

However, Proakis defines the system as a map from an input signal and a time coordinate to the output signal. If you want, the time coordinate itself can be understood as a second signal that just happens to contain the time index at each time index. Consequently, the time shift operator would act on this system definition by both shifting the proper signal and the time coordinate, so that the the system turns out to be time invariant.

So let me summarise: The system map $F(x,n)=x(n)u(n)$ is time invariant whereas the map $F(x)=x(n)u(n)$ is not time invariant.