Solution:
Let $r(k)$ be the remainder of $3^{k}$ upon division by $20$. In order to prove the claim, it is sufficient to show that for every natural number $k$ one has $r(k)<10$. It is clear that $3 \cdot r(k)$ and $r(k+1)$ have the same remainder upon division by $20$: one thus has $r(0)=1$, $r(1)=3$, $r(2)=9$, $r(3)=7$, $r(4)=r(0)=1$, $r(5)=r(1)=3$, $\ldots,$ and the sequence of the $r(k)$ turns out to be periodic with period $4$. It follows that the condition $r(k)<10$ is satisfied by every natural number $k$, since it is satisfied by the first $4$, and the claim is proved.

Alternative solution:
The units digit of a number of the form $3^{k}$ necessarily belongs to the set $\{1,3,7,9\}$, since such a number is odd and not divisible by $5$. We then proceed by induction on $k$: the claim is true in the case $k=0$; assuming that the tens digit of $3^{k}$ is $y$, the tens digit of $3^{k+1}$ coincides with the units digit of $3y$ or of $3y+2$, hence it is even by the induction hypothesis.