Why is this a non-holonomic constraint?

OP's example is clever. However in physics, the constraint function $$f$$ is often implicitly assumed to obey various regularity conditions, e.g. differentiability, which OP's example fails. For details, see e.g. this related Phys.SE post.

No. A system with $$n$$ degrees of freedom is said to be under a holonomic constraint if imposition of the constraint reduces the number of degrees of freedom by $$1$$ , to $$n-1$$. Every holonomic constraint can be captured by some relation of the form $$f(q_1...q_n,t)=0$$, but not every constraint that can be captured in this form is necessarily holonomic. You have provided an example.

When $$(x^2+y^2+z^2) = a^2$$ is the constraint satisfied, transforming to spherical coordinates (where one d.o.f is explicitly $$r= \sqrt{x^2+y^2+z^2}$$ ) allows us to describe the motion constrained on the sphere using $$\theta$$ and $$\phi$$, ie, reduction of the number of d.o.f from $$3$$ to $$2$$. Sometimes, it may not be straightforward to find this coordinate transform ; in such cases, we automate the process of describing the resultant $$n-1$$ dimensional motion by using the Lagrange Multiplier method.

So, calling a constraint on a system with $$n$$ d.o.f holonomic implies that imposing this constraint turns the system into one with n-1 d.o.f. This is by definition (see comments below)

What you have constructed indeed has the functional form described above, but it does not describe a situation with $$1$$ d.o.f less. The motion is very clearly still $$3$$-dimensional.