After you fixed the definition of $B(x,n)$, the Hausdorff property should be easy to check. Distinguish four cases: $x,y \in C_1$, $x \in C_1$ and $y \in C_2$ do [or do not] lie on the same ray emanating from the origin and $x,y \in C_2$.
Observe that the circle $C_1 \subset X$ carries its usual topology inherited from $\mathbb{R}^2$.
To check compactness of $X = C_1 \cup C_2$ I would use Alexander's subbase lemma (see also here for a proof), which tells us that it suffices to consider covers $\{U_{i}\}_{i \in I}$ consisting of sets from the subbase $\bigcup_{x \in X} \mathcal{B}_{x}$ of the topology of $X$.
Since $C_1$ has its usual compact topology, we can find $x_{1},\ldots,x_{n} \in C_1$ and $U_{i_k} = B(x_{k},n_k)$ such that $C_1 \subset U_{i_1} \cup \cdots \cup U_{i_n}$. But then all of $C_1 \cup C_2$ is covered, except possibly the points $2x_{k}$. Since we started with a cover of $C_{1} \cup C_{2}$, we can find $V_{k}$ containing $2 x_{k}$ from the cover $\{U_{i}\}_{i \in I}$ we started with. Then $\{U_{i_1},\dots,U_{i_n},V_{1},\dots,V_{n}\}$ covers $C_1 \cup C_2$ and is a subcover of $\{U_{i}\}_{i \in I}$. Thus we conclude from Alexander's lemma that $X$ is compact.
As I mentioned in a comment, this space $X$ is often called Alexandrov's double circle and appears e.g. as example 3.1.26 in Engelking's General topology (here are two screen shots containing the relevant excerpt: part 1 and part 2). It was first published on pages 13–15 of Alexandroff and Urysohn's Mémoire sur les espaces topologique compacts, Verh. Nederl. Akad. Wetensch. Afd. Naturk. Sect. I 14 (1929),1-96.
Its most prominent features are that it is a first countable compact Hausdorff space that contains a discrete subset of cardinality continuum (the circle $C_2$). Thus $X$ can't be second countable and it isn't separable either. Moreover, $C_2$ is open but not a countable union of closed sets, so $X$ is normal but not perfectly normal.