Spectrum of the multiplication operator

If $T$ is a linear operator on a (complex) Banach space $X$, then the resolvent set $\rho(T)$ is the collection of all complex numbers $\lambda$ such that $\lambda - T$ is a bounded invertible map $X \to \operatorname{Dom}(T)$. An important aspect of linear operators on Banach spaces is that their resolvent set is an open subset of $\mathbb{C}$. In fact, you learn that for $\lambda \in \rho(T)$, then the resolvent map $\lambda \mapsto (\lambda - T)^{-1}$ is analytic in $\lambda$ (that is, can be expressed as a power series within a sufficiently small ball around $\lambda$). This applies to your problem since the resolvent $\rho(T)$ is, in fact, the compliment of the spectrum $\sigma(T)$, implying that $\sigma(T)$ is closed. In your case, if $\lambda \in \{ q(s) : s \in [0,1]\}$ then suppose $q(t) = \lambda$. Now, let $f \in B[0,1]$ and $g \in B[0,1]$ be related by $f(s) = g(s)$ for every $s \neq t$ and $g(t) \neq f(t)$. Then for every $s \in [0,1]$ you have $(\lambda -q(s))f(s) = (\lambda-q(s))g(s)$. This means that $\lambda \in \sigma(T_q)$ since $\lambda - T_q$ is not invertible. Since $\sigma(T_q) = \overline{(\sigma(T_q))}$ (since it is closed) this implies that $\overline{\{q(t):t \in [0,1]\}} \subset \sigma(T_q)$. However, if $\lambda \notin \overline{\{q(t):t \in [0,1]\}}$ then $\lambda - q(s) \neq 0$ for any $s \in [0,1]$. In fact, there is some $\delta > 0$ such that $\sup_{s\in [0,1]} \| \lambda - q(s) \| > \delta$. It is easy enough now to see that $(\lambda - T_q)^{-1}f(s) = (\lambda - q(s))^{-1} f(s)$ and that $\| (\lambda - T_q)^{-1}\| \leq \delta^{-1} < \infty$. This means that if $\lambda \notin \overline{\{q(t):t \in [0,1]\}}$ then $\lambda \in \rho(T_q)$ and you have $\sigma(T_q) \subset \overline{\{q(t):t \in [0,1]\}}$ from which your question follows.