The Riemann Zeta function can be represented by the following series, which has also proven to be equivalent to the infinite product over primes:
$$\zeta(s)=\sum_{n}\frac{1}{n^{s}}=\prod_{p}(1-\frac{1}{p^{s}})^{-1}$$
The above is valid for $\sigma>1$ where $s=\sigma+it$.
A very common statement about $\zeta(s)$ is that it is non-zero for $\sigma>1$. The justification is that each of the factors of the infinite product are (1) non-zero and (2) finite.
Question: The justification seems insufficient. What am I missing?
For example, the following infinite product also has factors which are non-zero and finite, and yet the product is widely agreed to be zero (worked out here).
$$ \prod_{n=2}^{\infty}\left(1-\frac{1}{n}\right) = 0$$
It is true that the Euler product shows that $\zeta(s) \neq 0$ for $\mathrm{Re} s > 1$, but the reason is not what you gave. Instead, it's because the product converges absolutely at all.
Note that we say that $\prod_n a_n$ converges if the partial products tend to a nonzero limit --- it is necessary both for the partial products to converge to a limit and for the limit to not be zero.
So the statement that $\zeta(s) \neq 0$ for $\mathrm{Re} s > 1$ follows from the Euler product representation is a statement about convergence of the Euler product.
I should note that there is a well-known classical result that if $$ \sum_{n \geq 1} \lvert b_n \rvert $$ converges, then the partial products of $$ \prod_{n \geq 1} (1 + b_n) $$ converge to a limit $L$ and $L = 0$ if and only if one of the factors is $0$. This applies to $\zeta(s)$.
If you know this result, then it is true that since $\sum \lvert n^{-s} \rvert$ converges for $\mathrm{Re} s > 1$, that the Euler product representation of $\zeta(s)$ also converges and is nonzero because each individual term is nonzero.