The answer to the question is negative. Körner in [Hausdorff dimension of sums of sets with themselves][1] and Schmeling-Shmerkin in [On the dimension of iterated sumsets][2] showed that for any increasing sequence $\alpha_1 < \alpha_2 < \ldots < 1$, there is a compact set $A$, s.t. for every $k$, we have $\dim_{H} k A = \alpha_k$, where $kA = A + A + \ldots + A$ --- Minkowski sum of the set with itself $k$ times, and $\dim_H$ is the Hausdorff dimension of the set. In particular, choosing any sequence such that $\dim_H kA < 1$ for all $k$, none of the sum sets $kA$ could contain an interval.

See also the recent paper [Dimension growth for iterated sumsets][3] for some conditions under which we have dimension growth $\dim_H (A + A) > \dim_H(A)$. Unfortunately, it seems that the bounds there still fall short from implying that for some $k$, $\dim_H k A = 1$, much less providing quantitative bound on $k$. 


  [1]: https://www.impan.pl/en/publishing-house/journals-and-series/studia-mathematica/all/188/3/90488/hausdorff-dimension-of-sums-of-sets-with-themselves
  [2]: https://arxiv.org/abs/0906.1537
  [3]: https://arxiv.org/abs/1802.03324