Interpret the graph through its defining signature: the time taken for any concentration to halve is identical everywhere on the curve. That invariance of $t_{1/2}$ is only possible when elimination rate scales linearly with drug present.
Set up the rate law. If $\frac{dC}{dt} = -kC$, integration gives $C = C_0 e^{-kt}$, an exponential decay. The half-life is $t_{1/2} = \frac{0.693}{k}$, a quantity that contains no $C_0$ term - hence it is dose-independent. A fixed fraction (not amount) leaves per unit time. This is exactly what most drugs at therapeutic concentrations obey, because their metabolising enzymes and transporters are far from saturated.
Were the enzymes saturated, a constant amount would be cleared each unit time (zero-order), the plot would be a straight descending line and $t_{1/2}$ would lengthen as concentration rose - the behaviour of ethanol and high-dose phenytoin. The curve described does the opposite.
\[\boxed{\text{Constant } t_{1/2},\ C = C_0 e^{-kt} \Rightarrow \text{First-order kinetics}}\]