= Solution
The <Mackey restriction formula> says, for an $H$-representation $V$,
$$
\operatorname{Res}_H^G\operatorname{Ind}_H^GV
\cong
\bigoplus_{x\in H\backslash G/H}
\operatorname{Ind}_{H\cap xHx^{-1}}^H
\operatorname{Res}_{H\cap xHx^{-1}}^{xHx^{-1}}({}^xV).
$$
<Frobenius reciprocity> says
$$
\langle\operatorname{Ind}_H^G\alpha,\beta\rangle_G
=
\langle\alpha,\operatorname{Res}_H^G\beta\rangle_H.
$$
Applying both with an irreducible character $\chi$ gives
$$
\left\langle
\operatorname{Ind}_H^G\chi,
\operatorname{Ind}_H^G\chi
\right\rangle_G
=
\sum_{x\in H\backslash G/H}
\left\langle
\operatorname{Res}_{H\cap xHx^{-1}}\chi,
\operatorname{Res}_{H\cap xHx^{-1}}({}^x\chi)
\right\rangle.
$$
The identity double coset contributes one, and all terms are nonnegative integers. Therefore <Mackey irreducibility criterion> says that $\operatorname{Ind}_H^G\chi$ is irreducible exactly when every term belonging to a nonidentity double coset is zero.
Now take $G=SL_2(k)$. Write
$$
U=\left\{
\begin{pmatrix}1&b\\0&1\end{pmatrix}:b\in k
\right\},
\qquad
T=\left\{
\begin{pmatrix}a&0\\0&a^{-1}\end{pmatrix}:a\in k^\times
\right\},
$$
so $B=T\ltimes U$. For $q\geq4$, commutators of $T$ with $U$ generate $U$, because conjugation sends $u(b)$ to $u(a^2b)$ and one can choose $a^2\ne1$. Every degree-one character of $B$ is therefore trivial on $U$ and has the form
$$
\boxed{
\chi_\theta
\begin{pmatrix}a&b\\0&a^{-1}\end{pmatrix}
=\theta(a)
},
$$
where $\theta:k^\times\to\mathbb C^\times$ is a multiplicative character.
The <Bruhat decomposition of SL2 over a finite field> has two double cosets,
$$
G=B\sqcup BwB,
\qquad
w=\begin{pmatrix}0&-1\\1&0\end{pmatrix},
$$
and $B\cap wBw^{-1}=T$. Conjugation by $w$ inverts $T$, so
$$
\chi_\theta^w|_T=\chi_{\theta^{-1}}|_T.
$$
The only nonidentity Mackey term vanishes exactly when these two one-dimensional characters differ. Consequently
$$
\boxed{
\operatorname{Ind}_B^G\chi_\theta
\text{ is irreducible}
\quad\Longleftrightarrow\quad
\theta\ne\theta^{-1}
\quad\Longleftrightarrow\quad
\theta^2\ne1
}.
$$
Thus the trivial character is excluded, as is the unique quadratic character when $q$ is odd. All other degree-one characters of $B$ induce irreducibly; the induced representations have degree $[G:B]=q+1$. This is the <irreducible principal series of finite SL2>.
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