You think you always need a basis because you do not know enough examples. Perhaps the definition of a vector space over $\mathbb R$ would make things clearer.
A set $V$ together with operations $+ : V \times V \to V$ and $\cdot : \mathbb R \times V \to V$ is called a vector space over $\mathbb R$ if they satisfy these properties :
\forall a,b,c \in V, \quad a+b = b+a, \quad a+(b+c) = (a+b)+c,
and there exists $0 \in V$ such that $\forall a \in V, a+0 = a$. Furthermore, for each $v \in V$, there exists another vector called $-v$ such that $v+(-v) = 0$. For the scalar multiplication, we require
1 \cdot v = v, \qquad r \cdot (s \cdot v) = (rs) \cdot v,
and distributivity from both sides :
r \cdot (v_1 + v_2) = r \cdot v_1 + r \cdot v_2, \qquad (r_1 + r_2) \cdot v = r_1 \cdot v + r_2 \cdot v.
If you have something like this, it's called a vector space. You don't need a basis to express its elements. For instance, if you take the space of all functions from $\mathbb R \to \mathbb R$, they can be added ($(f+g)(x) = f(x) + g(x)$ is the definition of the sum of two functions), they all have inverses (the zero of this space would be the $0$ function) and multiplication by scalar is defined as $(\lambda f)(x) = \lambda (f(x))$. The properties are all satisfied, hence you have a vector space.
Note that this example is infinite dimensional (in the simple sense that it doesn't admit a finite basis). In infinite dimension, the discussion of bases is less trivial than in finite dimension.
A number usually stands for an element of $\mathbb R$, but for some people a number can also mean a complex number (i.e. an element of $\mathbb C$), so you need to notice the context. A vector stands for an element of a vector space, so saying that something is a vector refers to the vector space it lives in.
Hope that helps,