Spectra of a multiplication operator on $C^0(\overline\Omega)$

Let $\Omega \subset \mathbb{C}$ open and bounded, $X = C^0(\overline{\Omega})$ (vector space of continuous complex-valued functions). For an $y \in X$ with $y(\overline{\Omega}) = \{\lambda \in \mathbb{C} : \vert \lambda \vert \leq 1\}$ let $Tx = y \cdot x$. Calculate the spectra $\sigma_r$, $\sigma_p$ and $\sigma_c$.

From $\Vert Tx \Vert = \Vert y \cdot x \Vert \leq \Vert x \Vert$ it follows that $\Vert T \Vert = 1$, so I know that $\sigma(T) \subseteq \overline{B_1(0)}$. From $((\lambda - T)x)(t) = (\lambda - y(t))x(t)$ it follows that $\sigma_p = \emptyset$. I guess I have to divide this into different cases now and solve them one after another?

Is this true so far? How do I continue? I'm quite new to spectra so please, don't be that hard on me. (English also ain't my first language, sorry again)

Edit: Okay thanks alot user108903!

-
I don't see why the point spectrum should be empty. For example, if $y(K)=\{0\}$ for some set $K$ with non-empty interior, then $T$ will kill the functions supported on $K$, and $0$ is an eigenvalue. –  user108903 Jan 7 at 23:23
Okay. So if I have $\mathring{\Omega}_\lambda = \emptyset$ for all $\Omega_\lambda = y^{-1}(\{\lambda\})$, then I don't have any eigenvalues and I got $\sigma_p = \emptyset$ and $\sigma_c = \{\lambda \in \mathbb{C} : \vert \lambda \vert \leq 1\}$. On the other hand, if I got $\mathring{\Omega}_\lambda \neq \emptyset$ for a $\lambda \in \{\lambda \in \mathbb{C} : \vert \lambda \vert \leq 1 \}$, I got $\sigma_p = \{\lambda\}$ and $\sigma_c = \{\lambda \in \mathbb{C} : \vert \lambda \vert \leq 1\}$ ? –  JohnDoe746 Jan 8 at 18:05
I haven't thought about whether the situation I gave is the only way eigenvalues can arise; I'll leave that to you, although I think it's probably correct. On the other hand, there may be more than one eigenvalue, so your second statement needs some amendment. –  user108903 Jan 8 at 20:26