Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$?

If you take any sequence $a_1,a_2,a_3,\ldots$ whose sum is $\log_b (1/2)$, then $b^{a_1}, b^{a_2}, b^{a_3},\ldots$ is a sequence whose product is $1/2$.

Later note: Notice that $\frac 1 2 + \frac 1 4 + \frac 1 8 + \frac 1 {16} + \cdots = 1$. If you multiply every term by $\log_b \frac 1 2$ then you get a series whose sum is $\log_b \frac 1 2$.

Still later note: If $b>1$, then $\log_b(1/2)<0$, and $b^{a_n}$ will be in $(0,1)$ if $a_n<0$.

$$x_n=2^{-1/2^n}\qquad (n\geqslant1)$$

Try to find a sequence such that $\frac{n+1}{2n}=\prod_{j=2}^nx_j$ (it will do the job). We have $x_2=3/4$ and $$x_{n+1}=\frac{\prod_{j=2}^{n+1}x_j}{\prod_{j=2}^nx_j}=\frac{n+2}{2(n+1)}\frac{2n}{n+1}=\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{(n+1)^2}<\frac{n^2+2n\color{red}{+1}}{(n+1)^2}=1.$$ So $x_n=\frac{n^2-1}{n^2}$ does the job.