= Solution
After the reduction in part (b), multiply by $S$. The new bottom-left entry is $r$, whose absolute value is strictly smaller than the previous $|c|$:
$$
S T^nX=\begin{pmatrix}-c&-d\\r&b+nd\end{pmatrix}.
$$
Repeated <Euclidean division> must terminate because the nonnegative integers $|c|$ decrease strictly until zero. Thus a word $W$ in $S,T$ and their inverses gives a <matrix> with bottom-left entry zero. Its determinant is one, so the diagonal entries are equal and are either both $1$ or both $-1$:
$$
WX=\begin{pmatrix}\varepsilon&b_0\\0&\varepsilon\end{pmatrix}=\varepsilon I\,T^{\varepsilon b_0},\qquad\varepsilon\in\{1,-1\}.
$$
Since $S^2=-I$, this final <matrix> belongs to $\langle S,T\rangle$. So does $W^{-1}$, proving
$$
\boxed{SL(2,\mathbb Z)=\langle S,T\rangle}.
$$
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