5. In the Cartesian coordinate system, points with both coordinates being integers are called "integer points". Let $k$ be a non-zero real number. If the intersection point of the two distinct lines $y=k x-\frac{1}{k}$ and $y=\frac{1}{k} x-1$ is an integer point, then the number of possible values of $k$ is ( ).
(A) 1
(B) 2
(C) 3
(D) More than 3
5. A.

From the problem, we have
$$
\left(k^{2}-1\right) x=1-k \text{. }
$$

If $k=1$, then the equations of the two lines are the same, which does not meet the problem's requirements.
If $k=-1$, then equation (1) has no solution.
If $k \neq \pm 1$, clearly, $x \neq 0, k=-\frac{1}{x}-1$.
Then $y=\frac{1}{k} x-1=-x-\frac{1}{x+1}$.
Since $x, y$ are both integers, $\frac{1}{x+1}$ must be an integer.
Thus, $x+1= \pm 1 \Rightarrow x=0$ (discard) or -2.
Therefore, $k=-\frac{1}{x}-1=-\frac{1}{2}$.