You do that by induction on $k$. For $k=0$, we have $[0,1]$.
Suppose that for $k$ it holds, then $A_k$ is the union of $2^k$ closed intervals, each of length $3^{-k}$. Then $A_{k+1}$ is the set generated by removing the (open) middle third of each of these intervals, the remainder is two (closed) intervals of exactly $\frac13$ of the length. So we have $2\cdot2^k$ closed intervals, each of length $\frac13\cdot3^{-k}=3^{-(k+1)}$, as wanted.