I found interesting results when I put in certain values for z into zeta(z). I believe my findings do not disprove the Riemann hypothesis, but I am not so sure.
When I input z = 1/3 + i into the function, I got a result with a positive real component and a negative imaginary component, which can be written as a - bi.
When I input z = 1/3 - i, I got a result with a positive real component and a positive imaginary component, which can be written as c + di.
For z = 2/5 - i/2, I got -f + gi.
For z = 2/5 + i/2, I got -h - li.
Because the Riemann Zeta Function is continuous, these results imply that there is a number in this range z for which zeta(z) has a real component of zero. In fact, there would be infinitely many of these numbers in an unbroken chain. It also implies that there is a number in this range z for which zeta(z) has a complex component of zero and there there are an infinitely many of these in a broken chain. Also, these two chains should meet with each other at some point. This seems like it would disprove the Riemann Hypothesis.
Yes, you can draw a curve separating $\{\frac13 + i, \frac13 - i\}$ from $\{\frac25 + \frac i2, \frac25 - \frac i2\}$ where the real component of the zeta function is $0$.
Yes, you can draw a curve separating $\{\frac13 + i, \frac25 + \frac i2\}$ from $\{\frac13 - i, \frac25 - \frac i2\}$ where the imaginanary component of the zeta function is $0$.
But you have no control over where these curves intersect; they do not have to intersect inside the trapezoid whose corners are at $\{\frac13 + i, \frac13 - i, \frac25 + \frac i2, \frac25 - \frac i2\}$, because the curves we drew may leave that trapezoid. So the zero you find may be a trivial zero, or it may be a nontrivial zero with real part $\frac12$, neither of which would disprove the Riemann hypothesis.
In fact, here is the picture we get if we actually draw these curves: in blue is the curve with real component $0$, and in orange is the curve with imaginary component $0$. (There's more going on outside the range $[-3,3] \times [-3,3]$, of course.) We see them intersecting once at $z=-2$, and we see them pretending to intersect at $z=1$. (But they actually don't, because the zeta function has a singularity at $1$, corresponding to the harmonic series $1 + \frac12 + \frac13 + \dots$.)