What isn't a vector space?

First off, a vector space needs to be over a field (in practice it's often the real numbers $\Bbb R$ or the complex numbers $\Bbb C$, although the rational numbers $\Bbb Q$ are also allowed, as are many others), by definition. Thus, for instance, the set of pairs of integers with the standard componentwise addition is not a vector space, even though it has an addition and a scalar multiplication (by integers) that fulfills all of the properties we ask of a vector space.

A vector space needs to contain $\vec 0$. Thus any subset of a vector space that doesn't, like $\Bbb R^2 \setminus \{\vec 0\}\subseteq \Bbb R^2$ with the standard vector operations is not a vector space. Similarily, a vector space needs to allow any scalar multiplication, including negative scalings, so the first quadrant of the plane (even including the coordinate axes and the origin) is not a vector space.

A more subtle example is the circle (with some chosen zero) where addition is done by adding distances along the circle from the chosen zero (equivalently by adding angles), and scalar multiplication is done by multiplying distances (angles). Here we get into trouble with scalar multiplication again, because the zero vector is simultaneously representing $360^\circ$, so what should $0.5$ multiplied by that vector be? $0^\circ$? $180^\circ$? It would be both at the same time, which is not good.

Vector spaces are not just a set! They are an abstract concept, involving a set $V$, a field $\mathbb{F}$, and operations \begin{align*} + &: V \times V \rightarrow V \\ \cdot &: \mathbb{F} \times V \rightarrow V, \end{align*} addition and scalar multiplication respectively, satisfying a bunch of axioms. There's a lot more at play here than the set $V$ itself. Sets that can be made into vector spaces with the right field and operations are extremely common, but it's much rarer to be a vector space if the set already comes with the field and operations.

For example, the set of positive numbers $(0, \infty)$ doesn't seem like it's a vector space, but with scalar field $\mathbb{R}$ and with the (non-standard) operations,

\begin{align*} \oplus &: (0, \infty) \times (0, \infty) \rightarrow (0, \infty) : (x, y) \mapsto xy \\ \odot &: \mathbb{R} \times (0, \infty) \rightarrow (0, \infty), (\lambda, x) \mapsto x^\lambda \end{align*}

it forms a vector space. Even the natural numbers could be defined to be a vector space over a finite field, or a countable field like $\mathbb{Q}$ (although the operations would look a little funky).

So, to answer your question, can I come up with a set that is definitely not a vector space? Yes. As it turns out, all finite fields have a cardinality that takes the form $q = p^m$, where $p$ is prime and $m \in \mathbb{N}$. As such, finite vector spaces over such a finite field, which must have some finite dimension $n \ge 0$, must have cardinality $q^n$. Therefore, a set with a number of elements not equal to a prime power $p^{mn}$ must not be a finite vector space under any operations. For example, a set with $6$ elements is definitely not a vector space!

I. the set of points $(x,y,z)\in \mathbb R^3$ satisfying $x+y+z=1$ is not a vector space, because $(0,0,0)$ isn't in it. However if you change the condition to $x+y+z=0$ then it is a vector space.

II. The set of all functions from $\mathbb R$ to $\mathbb R$ is a vector space but the subset consisting of those functions which only take positive values is not.