For the product to be $≤0$, one of the values $x+1$ and $x$ has to be $≥0$ and the other has to be $≤0$. Consider the case when $x+1\geq 0$ and $x\leq 0$.
$$x+1\geq 0$$ $$x\leq 0$$
The solution satisfying both inequalities is $x\in \left[-1,0\right]$.
$$x\in \begin{bmatrix}-1,0\end{bmatrix}$$
Consider the case when $x+1\leq 0$ and $x\geq 0$.
$$x\geq 0$$ $$x+1\leq 0$$
This is false for any $x$.
$$x\in \emptyset$$
The final solution is the union of the obtained solutions.
