Let $K$ be a non-empty compact subset of $\mathbb{C}$, the complex field. Does there exist an operator $A\in \mathscr{B}(C[0,1])$ such that $\sigma(A)=K$?
$A$ is a multiplication operator iff $K$ is non-empty compact, connected and locally connected set in $C$. As the Hahn-Mazurkiewicz theorem states that:
A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.
How about the other kinds of compact sets? Any help would be appreciated!
Yes. The reason is that $C[0,1]$ is isomorphic as a Banach space to $c_0(C[0,1])$, the $c_0$-sum of countably many copies of itself. Let $E=c_0(C[0,1])$ and denote by $Q_n$ the projection onto the $n^{{\rm th}}$ copy of $C[0,1]$ in $E$.
Let $K$ be a non-empty compact subset of the complex plane and let $(a_n)_{n=1}^\infty$ be a sequence whose entries are dense in $K$. Define an operator $T\colon E\to E$ such that for each $x\in Q_n(E)$ we have $$Tx = a_n x.$$ All $a_n$ ($n\in \mathbb{N}$) are eigenvalues of $T$. Moreover, they are dense in the spectrum, so $\sigma(T)=K$.