For any function $f$ holomorphic on some domain $D$, the integral $$ \frac{1}{2\pi i} \int_{\partial D} \frac{df}{f} $$ is a nonnegative integer, and counts the number of zeros of $f$ in $D$. So if you can compute this integral to good enough precision, you know the exact number of zeroes in $D$. By taking $D$ a small enough disc around a suspected zero, you have a proof that there is actually a zero there.
Moreover, if $z$ is a zero of $\zeta$ such that $0<\Re(z)<1$, then $1-z$, $\overline{z}$, $1-\overline{z}$, are also zeroes in the same band (because of symmetries satisfied by the Zeta function). If there were a zero $z$ with $0<\Re(z)<1$, $\Im(z)=a$ and $\Re(z) \ne 1/2$, then $1-\overline{z}$ would be another zero satisfying the same conditions. For every small enough rectangle $R_\epsilon$ with vertices $a \pm i\epsilon$ and $1+a \pm i\epsilon$, $$ \frac{1}{2\pi i} \int_{\partial R_\epsilon} \frac{df}{f} $$ would be equal to $2$, and not to $1$.
