Although this question is over 2 years old now, I think it's interesting to consider the solution, assuming that the original interpretation was incorrect, and that $u[n]$ represents the system input, not the unit step sequence. If so, then the first thing to recognize is that the system is just an accumulator. For a time-domain demonstration, simply make repeated substitutions into the difference equation.
$$\begin{array}{rcl}
y[n] & = & y[n-1] + x[n] \\
& = & y[n-2] + x[n-1] + x[n] \\
& = & y[n-3] + x[n-2] + x[n-1] + x[n] \\
& = & y[n-4] + x[n-3] + x[n-2] + x[n-1] + x[n] \\
& \vdots & \\
y[n] & = & \displaystyle \sum_{m=0}^\infty x[n-m]
\end{array}$$
You can also show this using the $z$-transform,
$$\begin{array}{rcl}
Y(z) & = & z^{-1} Y(z) + X(z) \\
H(z) = \dfrac{Y(z)}{X(z)} & = & \dfrac{1}{1 - x^{-1}} \\
h[n] & = & \mbox{unit step sequence} \\
\Rightarrow y[n] & = & \displaystyle \sum_{m=0}^\infty x[n-m].
\end{array}$$
At this point, one can use the standard techniques to test if the system is linear, time-invariant, causal and stable.
Linearity: Define two input signals $x_1[n]$ and $x_2[n]$, with corresponding outputs $y_1[n]$ and $y_2[n]$,
$$\begin{array}{rcl}
y_1[n] & = & \displaystyle \sum_{m=0}^\infty x_1[n-m] \\
y_2[n] & = & \displaystyle \sum_{m=0}^\infty x_2[n-m]
\end{array}$$
Now define a third input signal that is a linear combination of the others,
$$ x_3[n] = a x_1[n] + b x_2[n]. $$
The output in response to $x_3[n]$ is
$$\begin{array}{rcl}
y_3[n] & = & \displaystyle \sum_{m=0}^\infty x_3[n-m] \\
& = & \displaystyle \sum_{m=0}^\infty \left( a x_1[n-m] + b x_2[n-m] \right) \\
& = & a \displaystyle \sum_{m=0}^\infty x_1[n-m]
+ b \displaystyle \sum_{m=0}^\infty x_2[n-m] \\
& = & a y_1[n] + b y_2[n].
\end{array}$$
Therefore the system is linear.
Time-Invariance: Using $x_1[n]$ and $y_1[n]$ as an input-output pair, define the input
$$x_2[n] = x_1[n-N]$$.
The corresponding output is
$$ y_2[n] = \displaystyle \sum_{m=0}^\infty x_2[n-m]
= \displaystyle \sum_{m=0}^\infty x_1[n-N-m]
= y_1[n-N]. $$
Therefore the system is time-invariant.
Causality: The system is causal by inspection, as $y[n]$ depends on $x[n-m]$ only for $m \geq 0$.
Stability: Let $x[n]$ be the bounded input signal that is 1 for all $n$. the output $y[n]$ is the sum of all present and previous inputs, which will diverge to infinity for all $n$. Therefore the system is unstable.