= Solution
Apply part (a) with $\chi=\xi=1$. It gives $\dim\operatorname{End}_G(V(1))=1$, so every endomorphism is a scalar multiple of the identity. The <local Steinberg representation of GL2> in the supplied sequence is nonzero: $V(1)$ is the space of locally constant functions on $B\backslash G\cong\mathbb P^1(F)$, and the characteristic function of the compact-open affine set $\mathfrak o$ is not constant. Its class in the quotient by constants is therefore nonzero.
If the sequence split, $V(1)$ would be $1\oplus\operatorname{St}$ with both summands nonzero. Projection onto either summand would be a $G$-endomorphism. These two projections are linearly independent: restrict a proposed linear dependence first to one summand and then to the other. This contradicts the one-dimensional endomorphism space. Equivalently the projection onto the trivial summand would be a nonzero idempotent different from the identity, whereas a scalar idempotent can only be zero or the identity. The principle that <scalar endomorphisms obstruct a split extension> gives
$$
\boxed{0\longrightarrow1\longrightarrow V(1)\longrightarrow\operatorname{St}\longrightarrow0\text{ does not split}.}
$$
No assertion of irreducibility of $V(1)$ is needed; this representation has a proper invariant line of constants.
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