Consider the short exact sequence of $C^\ast$-algebras
$$ 0 \rightarrow C_0(0,1)\overset{\imath }{\rightarrow} C[0,1] \overset{\psi}{\rightarrow} \mathbb C \oplus \mathbb C \to 0, $$ where $\imath: C_0(0,1) \to C[0,1]$ is the inclusion and $\psi(f) = (f(0),f(1))$ for each $f \in C[0,1]$.
I'm trying to proof that this exact sequence does not split.
Supposing that it does split, exists $\lambda: \mathbb C \oplus \mathbb C \to C[0,1]$ an $\ast$-homomorphism such that $\psi \circ \lambda = id|_{\mathbb C \oplus \mathbb C}$. Then, given $a,b \in \mathbb C$, we have
$$ (a,b) = \psi \circ \lambda (a,b) = ([\lambda (a,b)](0), [\lambda (a,b)](1)). $$ Hence, $\lambda (a,b) : [0,1] \to \mathbb C$ is a continuous path from $a$ to $b$.
How can I get a contradiction?