Can a non-zero vector field have zero divergence and zero curl?

I don't see how. Curl and divergence are essentially "opposites" - essentially two "orthogonal" concepts. The entire field should be able to be broken into a curl component and a divergence component and if both are zero, the field must be zero.

I'm visualizing it like a vector in $\mathbb{R}^2$. A vector cannot have a zero $x$ component and a zero $y$ component and still be non-zero.

EDIT: Here's a slightly more formal formulation of my thoughts: The way I see it, the curl and divergence form a "basis" - they are essentially orthogonal vectors. So how can a non-zero vector not be in their span?

Please don't just give me a counterexample. Please explain why my logic is incorrect.

You've had some complex analysis, so you know what a harmonic function is. Take the gradient of any harmonic function. They also have harmonic functions in three dimensions, same example.

You said not to do that. Life is tough.

Two dimensional, we can take harmonic function $x^2-y^2,$ which is the real part of $(x+yi)^2,$ to get vector field $$ (2x, -2y). $$ This has divergence zero and "curl" (as used in Green's Theorem) zero. It really is the curl, we just write it as a scalar.

No more difficult in three dimensions, we may take function $x^2 + y^2 - 2 z^2,$ giving vector field $$ (2x, 2y,-4z). $$ Again, zero divergence and zero curl.

Since you asked not just for a counterexample but also for an explanation of what is wrong with your reasoning, I would like to focus on what aspects of your intuition are correct, and where (I think) they fall short of capturing the full situation.

First, you are not entirely wrong in thinking that

the curl and divergence... are essentially orthogonal vectors

More precisely, I would say that you have noticed an important and useful property of vector fields: any vector field $\vec{F}$ can be decomposed as a sum of two fields $\vec{F} = \vec{A} + \vec{B}$, where $\vec{A}$ is "irrotational" (i.e. $\operatorname{curl} \vec{A} = 0$) and $\vec{B}$ is "incompressible" (i.e. $\operatorname{div} \vec{B} = 0$). Another way to say this is that $\vec{A}$ gets all of $\vec{F}$'s divergence and none of its curl, while $\vec{B}$ gets all of $\vec{F}$'s curl and none of its divergence.

This decomposition is, in some respects, analogous to decomposing a vector into its $x$ and $y$ components. An even better analogy might be to a force exerted on an object moving along a circular track; such a force can be decomposed into a radial component (pointing outward) and a tangential component (pointing along the direction of motion). In many respects it is reasonable to think of the divergence and curl of a vector field as being sort of like the radial and tangential components (respectively) of a vector.

But note that I said "sort of like", not "exactly like". There is one crucial difference, and it gets to the heart of your question. The difference is this: the decomposition of $\vec{F}$ into curl-free and divergence-free components is not unique. That is, we might have two different decompositions $$\vec{F} = \vec{A}_1 + \vec{B}_1$$ and $$\vec{F} = \vec{A}_2 + \vec{B}_2,$$ where $\vec{A}_1$ and $\vec{A}_2$ are both curl-free, and $\vec{B}_1$ and $\vec{B}_2$ are both divergence-free. How can this be? Well, we would need $$\vec{A}_1 + \vec{B}_1 = \vec{A}_2 + \vec{B}_2$$ which is equivalent to $$\vec{A}_1 - \vec{A}_2 = \vec{B}_2 - \vec{B}_1$$ Notice that the left-hand side of this equation is curl-free, and the right-hand side is divergence-free. So in order to satisfy it, we would need to find a nonzero vector field $\vec{H}$ satisfying both $\operatorname{curl}\vec{H} = 0$ and $\operatorname{div}\vec{H} = 0$. This is precisely what you are asking about! If we had such a vector field $\vec H$, then from any decomposition $\vec{F} = \vec{A} + \vec{B}$ we could form another one, namely $$\vec{F} = (\vec{A} + \vec{H}) + (\vec{B} - \vec{H})$$

Several of the other answers have given you examples of vector fields $\vec{H}$ having the required property, so you know they exist, and that means that the decomposition of a vector field into irrotational and incompressible components is not unique; it is in precisely this respect that the analogy breaks down, because the decomposition of a vector into its radial and tangential components is unique.

To go back to the wording of your question:

The way I see it, the curl and divergence form a "basis"

I would say that this intuition is half-right. To have a basis, you need vectors that both span a space, and are linearly independent. The "all-curl" (divergence-free) vector fields and the "all-divergence" (curl-free) vector fields do "span" the set of vector fields, in the sense that any vector field can be written as a sum of fields of those two types. But they are not "independent" because we do not have a unique way to write a vector field as a sum.

Using geometric calculus--the calculus of clifford algebra--we can write any vector field $F$ in terms of its value on a boundary curve $\partial M$ and its divergence and curl within a region $M$.

$$iF(p) = \oint_{\partial M} G(p-p') \, d\ell' \, F(p') + \int_M G(p-p') \, dA' \, \nabla F|_{p'}$$

where $G(p) = p/2\pi p^2$ is the 2d Green's function for $\nabla$. If $\nabla \cdot F = 0$ and $\nabla \wedge F= 0$, then $\nabla F = 0$ everywhere, and the area integral goes to zero.

But the line integral still remains, and $F$ is totally determined by its values on that bounding curve. A holomorphic function is determined by its values on some closed curve, is it not? This is just the 2d vector version of that concept.

So you can see, there are three parts to any decompsition of a vector field: a divergence-full part that is curl-free, a curl-full part that is divergence-free, and a divergence and curl-free part from the closed line integral (one word for this in the geometric calculus literature is monogenic, which is used to distinguish from the weaker condition of being harmonic).

That «the curl and divergence form a basis» does not really mean anything. The curl and the divergence are operators acting on vector fields, and they do not form a basis in any sense.

The contemplation of any counterexample to your claim should provide ample food for thought...

One way your logic fails is that both curl and divergence are differentials of the field, and differentials don't "see" constant terms. And consequently, the simplest counterexample to your claim is a non-zero constant field: It has zero curl and zero divergence everywhere, yet it is nowhere zero.