To prove that the span of a set of vectors forms a subspace of a vector space one can use the subspace test theorem. Suppose $W = span\{ v_1,v_2, \dots v_k \}$ where "span" means the set of all linear combinations with coefficients from $\mathbb{R}$. Note $W \neq \emptyset$ as $0$ is a linear combination $0v_1+0v_2+ \cdots +0v_k=0$. Moreover, if $x,y \in W$ and $c \in \mathbb{R}$ then $x = x_1v_1+ \cdots x_kv_k$ and $y=y_1v_1+\cdots +y_kv_k$ for some real constants $x_i,y_j \in \mathbb{R}$. Consider then:
$ cx+y = c[x_1v_1+ \cdots x_kv_k]+y_1v_1+\cdots +y_kv_k = (cx_1+y_1)v_1+\cdots + (cx_k+y_k)v_k $
Thus $cx+y \in W$. It follows that the nonempty $W$ is closed under scalar multiplication and vector addition and by the subspace test we find $W$ is a subspace. This means $W$ is a vector space with respect to the operations of the vector space $V$ which contains $W$.
Now, you can take the redundant set $\{ 1, \cos^2 \theta, \sin^2 \theta \}$ as a spanning set for your subspace $W$, however this would not be a basis.
To find a basis you need to select linearly independent vectors whose span is $W$. You already pointed out $\sin^2 \theta+\cos^2 \theta=1$ in your post. Think about this. You can see how to write one of the vectors in $\{ 1, \cos^2 \theta, \sin^2 \theta \}$ as a linear combination of the remaining vectors. You have at least three obvious choices for the basis here. Hope this helps.