Let $X=C[0,1]$ and we define
$$T\colon X\to X\quad\text{as}\quad(Tf)(t)=g(t)f(t)\quad\text{for all}\quad t\in [0,1]$$ with $g\in C[0,1]$ fixed.
I must find the continuous spectrum.
Now, if $\lambda\in g([0,1])$, then $\lambda=g(t_0)$ for some $t_0\in [0,1]$. As a consequence we have for all $h\in\overline{\text{Ran}(T-\lambda I)}\subset X$ that $h(t_0)=0$
Question. Why this implies that $\overline{\text{Ran}(T-\lambda I)}$ cannot be whole space $X$?
I am assuming you are working with the uniform norm and so since there are continuous functions in $C([0,1])$ such that $g(t_0)>\epsilon$ you cannnot have that the image is dense, since these functions will never be in it's closure.Now let's see why this statement is true, keep in mind that we are working with the uniform metric as our norm and so for any two functions $f,g $ we have $||f-g||=sup_{x\in [0,1]}|f(x)-g(x)|. $ Now take an arbitary function $f$ such that $f(t_0)\geq \epsilon$ we know this is possible because we can construct such continuous function. Now for any $h\in Im(T_{\lambda})$ we will have that $h(t_0)=0$ since $\lambda=g(t_0)$, and so we will have that $||f-h||=sup_{x\in [0,1]}|f(x)-h(x)|\geq |f(t_0)-h(t_0)|=|f(t_0)|\geq\epsilon $ and so we will have that $B(f,\epsilon)\cap Im(T_{\lambda})=\emptyset$ and so we have that $f\notin cl(Im(T_{\lambda}))$ and so the image cannot be dense.
But this means that the residual spectrum is equal to the range of $g$, if you assume that $\lambda$ is not there you can show that the inverse operator is bounded since the $||(T_{\lambda}f(t))^{-1}||=||\frac{f(t)}{g(t)-\lambda}||\leq ||f||M$ where M is the maximum of the continuous function $\frac{1}{g(t)-\lambda}$ in $[0,1]$ and so $\sigma_c(T)=\emptyset$.
And assuming that $g$ is not constant you have $\sigma_p(T)=\emptyset$,and you will have that $\sigma(T)=range(g(t)).$With this result you can actually see an interesting result that for any set of the form $[a,b]$ we can find an operator $T$ such that $\sigma(T)=[a,b]$, we just need to find the appropriate function for wich to consider in the multiplication operator.