= Solution
The strict statement as printed needs the qualification \b[if $F\ne0$]. Part (a) already shows $F^2=0$ in the present situation, and the next part proves $F=0$; an unconditional $F^2<0$ would contradict that conclusion.
To prove the needed conditional statement, suppose $F\ne0$. A <fixed part> has $h^0(X,F)=1$. Indeed, a different $F'\in|F|$ has smaller multiplicity along some component of $F$; otherwise $F'-F$ would be a nonzero effective linearly trivial divisor. Adding a member of the <movable part> avoiding that component contradicts fixedness.
The <Hodge index theorem for algebraic surfaces>, together with $D\cdot F=0$ and $D^2=0$, gives $F^2\leq0$. If equality held, <Riemann–Roch theorem for algebraic surfaces> would give $\chi(X,F)=2$, and <Serre duality> would give $h^2(X,F)=h^0(X,-F)=0$. Hence $h^0(X,F)=2+h^1(X,F)\geq2$, a contradiction. Therefore
$$
\boxed{F\ne0\ \Longrightarrow\ F^2<0.}
$$
This is the <fixed-part elimination on a K3 surface> argument.
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