Solving recurrences by substitution
Assuming for simplicity that $n$ is even, if we guess that $T(n) \leq C^2n$ then we can prove our guess by induction (for powers of 2) for an appropriate value of $C$ since
$$
T(n) = 2T(n/2) + 1 \leq C(n/2)^2 + 1 = \frac{C}{4} n^2 + 1
$$
is smaller than $Cn^2$ as long as $C \geq 4/3$ (assuming $n \ge 1$).