Linearity, Causality and Stability of a System
According to the interpratation (that belongs to you) that output $y[n]$ is given by the recursion relation $$y[n] = y[n-1] + u[n]$$ no matter what the input $x[n]$ is, then the following conclusions can be made about the linearity, time-invariance, stability and causality of the system.

First let's solve the output for $y[n]$

i.For $n&lt;0$, since $u[n]=0$, we have $$y[n] = y[n-1]$$ which implies that $y[-1]=y[-2]=...y[-N]...= K$ for all $n &lt; 0$

ii.For $n\geq0$, since $u[n]=1$, then we have
$$y[n] = y[n-1] + 1$$ which can be recursively solved to get

\begin{align}
y[0] &amp;= y[-1] + 1 = K +1 \\
y[1] &amp;= y[0] + 1 = (K+1) +1 = K+2 \\
y[2] &amp;= y[1] + 1 = K +3 \\
&amp;...\\
y[n] &amp;= y[n-1] + 1 = K +(n+1) \\
\end{align}

Combining them yields: $$\boxed{y[n] = K + (n+1) u[n] }$$ for all $n$ as th fixed-unique output of the system for all inputs $x[n]$

Then we analyse the system for its properties:


The system is non-linear:
It's so evident since the system has a fixed, unique output for every input
it's a non-linear system as it cannot satisfy $$\mathcal{T}\{a x_1[n]+ b x_2[n] \} = a \mathcal{T}\{x_1[n] \} + b\mathcal{T}\{x_2[n]\} $$
The system is time-varying:
Again shifting the input have no effect on the output (which is fixed) so it cannot satisfy $$ y[n-d] = \mathcal{T}\{x[n-d] \}$$
NOTE: Does this make any sense? The system is already a fixed one. It has a fixed output for every input. Hence effectively nothing varies in its output. Nevertheless I called it as time-varying, because evidently it cannot saitsfy the time-invariance constraint.
The system is causal: it only depends on the past value of the output and it does not depend on any value of the input so it does not violate causality.
The system is unstable: as for a bounded input $x[n] = \delta[n]$ the output is $y[n] = K + (n+1) u[n]$ which grows unbounded as $n$ goes to infinity.
The system is not invertible.
The system can be implemented with and without any memory requirement.