Let $T$ be a bounded operator on a Hilbert space $H$. For any such thing we have a holomorphic functional calculus (by Cauchy residue formula), which is a continuous $\mathbb{C}$-algebra homomorphism:
$$Hol(\sigma(T)) \to \mathcal{B}(H)$$
Where we denote by $Hol(\sigma(T))$ the algebra of germs of holomorphic functions defined in some neighborhood $\sigma(T) \subset U\subset \mathbb{C}$ (endowed with the topology of uniform convergence on compact subsets). For some purposes it is desirable to upgrade this calculus to a larger class of functions.
Recall that the functional calculus extends the polynomial functional calculus which itself is a restriction of the natural "evaluation" $*$-homomorphism:
$$\mathbb{C}<z,\bar z> \to \mathcal{B}(H)$$
Where the LHS is the free $*$-algebra generated by $z$ and $\bar z$ with involution $z \mapsto \bar z$ and the homomorphism takes $z \mapsto T$ and $\bar z \mapsto T^*$
Consider now the embedding $\mathbb{C}[z,\bar z] \to C_0(\mathbb{C})$ of polynomials into complex valued continuous functions. This is naturally a homormophism of $*$-algebras (the involution on the LHS being $z \mapsto \bar z$). The RHS is moreover naturally a $C^*$-algebra and the embedding is dense in the topology determined by the $C^*$-norm (Stone-Weirstrass theorem).
In order for the evaluation calculus $\mathbb{C}<z,\bar z> \to \mathcal{B}(H)$ to upgrade to a continuous functional calculus it must first factor through $\mathbb{C}[z,\bar z]$ (this happens presicely when $T$ is normal). Suppose this happens, then we could say by the above paragraph that the polynomial functional calculus upgrades to a continuous functional calculus iff its continuous w.r.t. the topology on $\mathbb{C}[z,\bar z]$ inherited from the enbedding into $C_0(\mathbb{C})$. But since the spectral radius determines the norm of a $C^*$-algebra (which determines its topology) any homomorphism of $C^*$-algebras is automatically continuous.
We conclude that $T$ has a continuous functional calculus iff its evaluation homomorphism $\mathbb{C}<z,\bar z> \to \mathcal{B}(H)$ factors through $\mathbb{C}[z,\bar z]$. This is true iff $T$ is a normal operator. We conclude the following:
A bounded operator $T$ on a Hilbert space has a continuous functional calculus iff it is normal.