Let $B$ be the Banach space of bounded complex functions on $[0,1]$ with sup-norm. For $q \in B$, define the (multiplication) operator $T_q : B\rightarrow B$ by $(M_q f)(t) = q(t)f(t)$.
How do you show that $\sigma(T_q) = \overline{\{q(t) : t \in [0,1]\}}$ for each $q \in B$?.
If $\lambda=q(s)$ for some $s\in[0,1]$, then $T_q 1_s=q 1_s=q(s) 1_s=\lambda 1_s$. Thus, $\lambda\in \sigma(T_q)$. As the spectrum of an operator is closed, it follows that $\overline{\{q(t)\mid t\in[0,1]\}}\subset \sigma(T_q)$.
If $\lambda\notin\overline{\{q(t)\mid t\in[0,1]\}}$, let $\phi=\frac 1{q-\lambda}$. Since $\lambda\notin\overline{\{q(t)\mid t\in[0,1]\}}$, we have $d:=\inf\{|\lambda-q(t)|\colon t\in[0,1]\}>0$. Thus, $|\phi(t)|\leq \frac 1 d$ for all $t\in [0,1]$. Hence $T_\phi$ is bounded, and it is easy to see that $(T_q-\lambda)M_\phi=T_\phi(T_q-\lambda)=\mathrm{id}$. This settles $\sigma(T_q)\subset \overline{\{q(t)\mid t\in[0,1]\}}$.