Solution:

We can form at most $k^r$ distinct sums, so $k^r$ must be $\geq n$.

Now consider $A_1 = \{0, 1, 2, \ldots, k-1\}$, $A_2 = \{0, k, 2k, \ldots, (k-1)k\}$, $A_3 = \{0, k^2, 2k^2, \ldots, (k-1)k^2\}, \ldots, A_r = \{0, k^{r-1}, 2k^{r-1}, \ldots, (k-1)k^{r-1}\}$.

Then for any non-negative integer $m < k^r$, we can write $m$ with $r$ digits in base $k$ (using leading zeros as necessary) and hence as a sum of one element from each $A_i$.

This subset works for $(k-1)k^{r-1} < n \leq k^r$. For smaller $n$ above $(k-1)^r$ we cannot use all the elements given above, but we do not need them, so we just replace the elements which are too large by arbitrary elements under $n$.

For example, suppose $n = 17$, $r = 4$. We need $k = 3$. So we form $A_1 = \{0, 1, 2\}$, $A_2 = \{0, 3, 6\}$, $A_3 = \{0, 9, 18\}$, $A_4 = \{0, 27, 54\}$. Now $18, 27, 54$ are unnecessary, so we pad out $A_3$ and $A_4$ with other elements. We could take $A_3 = \{0, 1, 9\}$, $A_4 = \{0, 1, 2\}$.