How can I prove that this infinite product equals $0$?

An approach that does not rely on combinatorics or factorials:

$$ \left(\prod_{n=1}^\infty\frac{2n-1}{2n}\right)\left(\prod_{n=1}^\infty\frac{2n}{2n+1}\right)=\prod_{n=1}^\infty\frac{2n-1}{2n+1}=\lim_{n\to\infty} \frac1{2n+1} = 0 $$ where the $\lim$ expression arises due to telescoping.

But both of the terms on the left are positive, and it is easy to confirm that $$ \frac{2n-1}{2n}<\frac{2n}{2n+1} $$ and thus $$ a=\left(\prod_{n=1}^\infty\frac{2n-1}{2n}\right)\leq\left(\prod_{n=1}^\infty\frac{2n}{2n+1}\right)=b $$ where equality occurs if both are zero. Therefore, as $0\leq a\leq b$ and $ab=0$, we must conclude that $a=0$.

You may notice that: $$ p_n = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right) = \frac{1}{4^n}\binom{2n}{n} \tag{1}$$ has a square given by: $$ p_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)=\frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2} $$ In particular: $$ p_n^2 \leq \frac{1}{4n}\exp\sum_{k=2}^{n}\left(\frac{1}{4(k-1)}-\frac{1}{4k}\right) \leq \frac{e^{1/4}}{4n}\tag{3} $$ and: $$ p_n \leq \frac{C}{\sqrt{n}}\tag{4} $$ with $C=\frac{1}{2}e^{1/8}$.

Apply Stirling's approximation $$n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ to $$\frac{(2n)!}{2^{2n}(n!)^2}.$$ Simplifying gives $$\frac{1}{\sqrt{n\pi}}\to 0.$$