Let the Banach space $X=C[0,1]$, and consider the transformation $T:X\to X$ defined by $Tx=vx$ for a fixed $v\in X$.
I would like to find the spectral values of $T$, denoted $\sigma(T)$. To do this we note that when $R_\lambda(T)=T_\lambda^{-1}=(T-\lambda I)^{-1}$ exists, it is bounded and defined on the space $X$, so the spectral values are the values of $\lambda$ for which $R_\lambda(T)$ doesn't exist. Because $T_\lambda$ is linear, it's inverse exists if and only if $T_\lambda x=0$ implies $x=0$.
I know that that the solution is $\sigma(T)=\text{Range}(v)$. It seems to me that this is because for a $\lambda\in \text{Range}(v)$, there is a $t_0$ such that $v(t)=\lambda$, and thus at this $t_0$, $(v-\lambda)=0$.
My question is: why is it sufficient to find a single point $t$ such that $v(t)-\lambda=0$ to make our function transformation non-invertible? Is there some way to construct a non-trivial $x\in X$ such that $(v-\lambda)x=0$?
Edit: The answer is that our transformation is not surjective, and thus not invertible. It is not surjective because $Y=T_\lambda X \subsetneq X$ because for all $y\in Y,\ y(t_0)=0$.
I am not sure about this answer - maybe someone with higher knowledge can doublecheck this.
It seems like you know the answer, but you don't the reasoning for it. To determine spectral values, normally you compute the resolvent set - which, as you said, is the set of complex values such that $R_\lambda = (T-\lambda I)^{-1}$ exists and is bounded in the whole space $C[0,1]$.
Well, if you want to find values of $\lambda \in \mathbb C$ such that $R_\lambda$ exists, it suffices to solve the equation below in order to $f$. $$ ((T-\lambda I)f)(x) = g(x), \quad \forall x \in [0,1]. $$
If you do this, you will obtain the following result: $$ f(x) = \frac{g(x)}{v(x)-\lambda} \Leftrightarrow f(x) = ((T-\lambda I)^{-1}g)(x) \quad \forall x \in [0,1].$$ Obviously, such operation is valid iff $v(x) \neq \lambda.$ This means that $R_\lambda$ exists for every $\lambda$ s.t. $v(x) \neq \lambda, \forall x \in [0,1].$
It is left to prove that $(T-\lambda I)^{-1}$ is bounded but I think this was your main doubt.
EDIT.
To clarify the OP doubts, everything I did to support my claim was to solve the equation. I will do it explicity below:
$$ ((T-\lambda I)f)(x) = (Tf)(x) - ((\lambda I)f)(x) = v(x)f(x) - \lambda f(x), \quad \forall x \in [0,1].$$ With this in mind, it follows that $$ ((T-\lambda I)f)(x) = g(x) \Leftrightarrow v(x)f(x) - \lambda f(x) = g(x) \Leftrightarrow f(x) = \frac{g(x)}{v(x)-\lambda}, \forall x \in [0,1].$$ Then, the conclusions follow.