Induction problem

It is enough to prove the more general statement
\[(*)          (1-𝑥_1)⋯(1-𝑥_n)≥1-𝑥_1-x_2- \dots -𝑥_𝑛.\]
Indeed if (*) holds, then $x_1+x_2+\dots x_n \leq \frac{1}{3}$ implies
\[(1-𝑥_1)⋯(1-𝑥_n)≥1-𝑥_1-x_2- \dots -𝑥_𝑛 \geq 1-\frac{1}{3}=\frac{2}{3},\]
which is what you want to prove.

We prove $(*)$ by induction. For $n=1$ $(*)$ becomes 
\[(1-𝑥_1) \geq (1-𝑥_1)\]
which is clearly true. 

Assume $(*)$ holds for $n$. We show that it also holds for $n+1$. If we now multiply the inequality in $(*)$ by $(1-𝑥_{𝑛+1})≥0$ we obtain

\[(1-𝑥_1) \dots (1-𝑥_𝑛)(1-𝑥_{𝑛+1})≥(1-𝑥_1- \dots -𝑥_𝑛)(1-𝑥_{𝑛+1})\]
\[=(1-𝑥_1-\dots-𝑥_𝑛)-𝑥_{𝑛+1}+𝑥_{𝑛+1}(𝑥_1+\dots+𝑥_𝑛)≥1-𝑥1- \dots- 𝑥_{𝑛+1},\] where the last inequality follows from the fact that $𝑥_𝑖 \geq 0$ which implies $𝑥_{𝑛+1}(𝑥_1+\dots+𝑥_𝑛) \geq 0$