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Is it known whether or not $\zeta(\sigma)\neq 0$ for $0<\sigma<1$, where $\zeta(\sigma)$ is the Riemann zeta function.

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    $$\eta(s) = \sum_{n = 1}^{\infty} \frac{(-1)^{n-1}}{n^s} = \bigl( 1 - 2^{1-s}\bigr)\zeta(s)$$ By Leibniz, $\eta(\sigma) > 0$ for $0 < \sigma$, and since $1 - 2^{1-\sigma} < 0$ for $0 < \sigma < 1$, the result follows. – 2017-01-16
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    @DanielFischer That's cleaner than what I have in mind. Very nice – 2017-01-16

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Yes, this is known.

Often one see this while working through an analytic proof of the prime number theorem with error term. In particular, the real line segment $(0,1)$ is contained in the so-called zero free region. You should be able to find a proof by looking at an analytic number theory text that includes a proof of the prime number theorem, and by looking for the section including the zero free region.

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    I'm not sure I would agree with this. The zero-free region is usually meant to mean the region $\sigma > 1 - c/\log(|t| +3)$, or something like that (though of course $\zeta(s)$ can be shown to be nonzero in many other regions of the complex plane). The fact that $\zeta(\sigma) < 0$ for $0 < \sigma < 1$ is something completely separate. An easy proof comes from the identity $\zeta(s) = \frac{s}{s - 1} - s\int_{1}^{\infty} \frac{\{x\}}{x^s} \, \frac{dx}{x}$, valid for $\Re(s) > 0$ with $s \neq 1$, which follows easily from partial summation. – 2017-01-16
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The real zeroes of the zeta function are well known; they are $-2, -4, -6, \ldots$ and no others. None of these are in $(0,1)$.