Interesting question.
For $k>1$, the chevyshev inequality guarantees that we will always have a non-trivial proportion of data points left. If we do this procedure a finite number of times, then some data points will always be left.
We could talk about taking the limit to infinity. At which point, simply show that if there is less than $N=k^2$ points of data, since $\frac{N-1}{N} < 1 - \frac{1}{k^2}$, this implies that we cannot lose any data points in this procedure.
For $k=1$, note that the uniform distribution on 2 points will result in both points lying exactly 1 SD away. Arguably, you'd want to throw this out, and hence we are left with 0 points.
For $k<1$, the uniform distribution on 2 points will result in both points being thrown out. Of course, several other distributions work too, esp those that are very heavy in the tail ends. Note that chevyshev becomes trivial, and hence doesn't apply in this case.