Consider $C$ the set of circles formed by the two functions $x(\theta), y(\theta)$ which verifies:
$$C =\lbrace \forall r \in \mathbb{R}, \theta \in [0,2\pi]; x(\theta)=r\cos(\theta), y(\theta)=r\sin(\theta) \rbrace $$
Prove that $C$ is a subspace. Any hints or help would be greatly appreciated! thanks.
OK. First, the space we're looking at is the space of functions, $$ E = \Bbb R^{\Bbb R} $$ That's got a well-defined addition and scalar multiplication, and a zero vector, namely the function $$ z: \Bbb R \to \Bbb R : x \mapsto 0. $$
The set $C$ is a set of functions. I suspect that what's meant is that $$ C =\lbrace C_r \mid r \in \mathbb{R} \rbrace $$ where $$ C_r : \Bbb R \to \Bbb R^2 : \theta \mapsto (r \cos \theta, r \sin \theta) $$ i.e., each $C_r$ is a function, and $C$ is the set of all such functions.
The difficulty here is that $C_r$ is not an element of $E$, because it's a function from $\Bbb R$ to $\Bbb R^2$ rather than to $\Bbb R$. So assuming that you missed something, I'm going to guess that $E$ is really $$ E = (\Bbb R^2)^{\Bbb R}. $$ Once again, addition of functions is defined (via addition of vectors in the codomain) and scalar multiplication arises similarly.
Now the zero function is $$ z: \Bbb R \to \Bbb R : x \mapsto (0, 0). $$
OK. So is $C$ a subspace of this particular (new) vector space $E$? Yes.
For if we take any two elements $$ C_r $$ and $$ C_s$ $$ of $C$, where $s$ and $r$ are any real numbers, and a number $b$, we can show that $$ C_r + b C_s \in C $$
In fact, it's really quite easy: I claim that $$ C_r + b C_s = C_{r + bs}. $$ We can check that. They are two functions; they have the same domain and same codomain, so all that remains to check is that for every $\theta \in \Bbb R$, we have $$ (C_r + b C_s)(\theta) = C_{r + bs}(\theta). $$ Let's do that. \begin{align} (C_r + b C_s)(\theta) &= C_r(\theta) + (b C_s)(\theta) & \text{ definition of addition of functions}\\ &= C_r(\theta) + b (C_s(\theta)) & \text{ definition of scalar multiplication of functions}\\ &= (r \cos \theta, r \sin \theta) + b (s \cos \theta , s \sin \theta) & \text{ definition of $C_u$ for any $u$, given above}\\ &= (r \cos \theta, r \sin \theta) + (bs \cos \theta , bs \sin \theta) & \text{ definition of scalar multiplication in $\Bbb R^2$}\\ &= ((r+bs) \cos \theta, ((r+bs) \sin \theta) & \text{ definition of vector addition in $\Bbb R^2$}\\ &= C_{r + bs}(\theta) & \text{definition of $C_u$, for $u = r + bs$}. \end{align}
And we're done.