So I am thinking, if this graph I have made is incorrect, since the distances between \$i-i_2\$ and \$i_1-i\$ should be the same, as depth of focus is symmetrical.
(Emphasis mine) Your assumption that depth of focus is symmetrical is incorrect. Your diagram is correct; it follows from the principle of the thin lens formula.
And if you follow through the math, you'll note that the near depth of field \$o - o_1\$ is always smaller than the far depth of field \$o_2 - o\$. As you approach highest magnification (i.e., \$i\$ becomes very large, and \$o\$ approaches \$f_+\$, the ratio of the far depth of field to near depth of field approaches unity (from above):
$$
\lim_{o\to f}\left(\frac{o_2-o}{o-o_1}\right) = 1^+\,.
$$
So similarly, the ratio of near and far depths of focus would be
$$
\lim_{i\to \inf}\left(\frac{i-i_2}{i_1-i}\right) = 0\,.
$$
And focusing far away, the depth of focus ratio limit (near/far) would look similar to the depth of field limits for high magnification:
$$
\lim_{i\to f}\left(\frac{i-i_2}{i_1-i}\right) = 1^-\,.
$$
While your diagram exactly illustrates the geometry, the proportions aren't realistic. Of course, that's okay, it's for derivation or illustration purposes.
But let's imagine what happens in a more realistically-proportioned system: your circle of confusion \$C\$ is many times smaller, probably less than a pixel of your drawing size. So if you were to zoom into the diagram with a more realistic \$C\$, you'd see the marginal rays for \$i\$, \$i_1\$, and \$i_2\$ appear very nearly (but not mathematically exaactly) parallel to each other. And if you draw those marginal rays as parallel, then in the limit, yes, the near and far depths of focus would be symmetrical.
