A hyperelliptic curve is never a complete intersection

The reason is that a complete intersection has its canonical class forced to be a multiple of the hyperplane class, and a positive multiple of the hyperplane class is very ample.

Adjunction on the complete intersection gives \[ \omega_C \cong \OO_C(a), \qquad a = \sum_{i=1}^{n-1} d_i - n - 1 , \] and \(\deg \OO_C(1) = \prod d_i =: d > 0\). So \(2g-2 = ad\), and the hypothesis \(g \geq 2\) forces \(a \geq 1\).

Now \(\OO_{\PP^n}(a)\) is very ample for \(a \geq 1\), so the linear system it cuts on \(C\) separates points and tangent vectors. That cut system is a subsystem of the complete system \(\abs{\OO_C(a)} = \abs{K_C}\), and a linear system containing a very ample subsystem is itself very ample. Hence \(K_C\) is very ample, so the canonical map is an embedding, so \(C\) is not hyperelliptic. (Complete intersections are projectively normal, so in fact the cut system is all of \(\abs{K_C}\): the canonical embedding of \(C\) is the composite of \(C \subseteq \PP^n\) with the \(a\)-uple embedding. The weaker subsystem statement is already enough for the conclusion.)

The contrapositive is what the exercise asks for: a hyperelliptic curve embeds in projective space in many ways, but never as a complete intersection in any \(\PP^n\). Two checks against curves one can name. Genus \(2\) is entirely hyperelliptic, and such a curve embeds in \(\PP^3\) by a divisor of degree \(5\); it cannot be a complete intersection there, and indeed the only factorization \(5 = d_1 d_2\) is \(1 \cdot 5\), which would place the curve in a plane as a quintic of genus \(6\). Genus \(4\) runs the other way: the non-hyperelliptic canonical curve is the intersection of a quadric and a cubic in \(\PP^3\), with \(a = 2+3-3-1 = 1\) and \(\omega_C \cong \OO_C(1)\), exactly as the formula predicts for a canonically embedded complete intersection.