Current results are able to yield such results. Depending on how generous one is regarding what $X$ is. If it is just the optimal value can be calculated *exactly* this will work for many more $k$ and if one is happy with an explicit bound for all $k$.

For example Dusart showed that
$$
\frac{x}{\log x - 1} \le \pi(x) \le \frac{x}{\log x - 1.1}
$$
for $x\ge 60184$.
Now for some $k$, write $y=kx$.
Then, if the upper bound for $kx=y$ is smaller than the lower bound for $(k+1)x = (1+1/k)y$, that is
$$
\frac{y}{\log y - 1.1} \lt \frac{y(1+ 1/k)}{\log( y (1+1/k) )- 1}
$$
one has a prime between $kx$ and $(k+1)x$, since then $\pi(kx) \lt \pi((k+1)x)$.

One can check that this inequality holds for (up to potential error in my calculation)
$$
y \ge 10 e^{0.1 k}.
$$

So, for $x \ge \max \lbrace 10 e^{0.1 k}/k , 60184/k \rbrace $ one always has a prime between $kx$ and $(k+1)x$.

While this grows exponential in $k$, the growth is such that it is well feasible to check 'everything' up to the bound to get an optimal $X$ for not too large $k$. And, one always has an explict value.

This proof is of course not elementary (the non-elementariness being hidden in Dusart's result) and is an application of the PNT in some sense. But what this is meant to show is that for a result around this to be interesting it seems necessary either to be better (and one could still optimize this here) than this *or* the proof would have to be interesting (or both). [What an interesting proof is is of course a bit subjective.]

Acta Arithmetica52(1989), 307-337. $\endgroup$ – Noam D. Elkies Aug 30 '13 at 23:23