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The embedding line bundle has at least four independent global sections. If it is nonspecial, the Riemann-Roch theorem gives $h^0=6-g$, so $g\le 2$. If it is special, the Clifford inequality for curves gives $h^0\le 3$, a contradiction. The smooth projective curve need not span its full ambient projective space.

Let $\mathcal{L}={\mathcal{O}}_C(1)$. Its degree is five. Because the linear span of $C$ has dimension at least three, restrictions of linear forms supply at least four independent global sections, soIf $h^1(C,\mathcal{L})=0$, the Riemann-Roch theorem gives $h^0(\mathcal{L})=6-g$, and therefore $g\le 2$ immediately.

If instead $\mathcal{L}$ is special, the Clifford inequality for curves givesso its integer dimension is at most three, a contradiction. Here is a short proof of the required Clifford inequality for curves, so the bound need not be assumed. Put $r=h^0(\mathcal{L})>0$ and $s=h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$. At a smooth geometric point, choose bases of these two section spaces with strictly increasing orders of vanishing $a_1<\cdots <a_r$ and $b_1<\cdots <b_s$. Such bases exist because each successive order-of-vanishing quotient has dimension at most one. The products of the first basis with the first element of the second basis, followed by products of the last element of the first basis with the remaining elements of the second, have ordersThey are linearly independent global sections of ${\omega }_C$, proving the product dimension bound for sections on a curve. Thus $r+s-1\le h^0({\omega }_C)=g$. Combining this with Riemann-Roch theorem, $r-s=\mathrm{deg}\mathcal{L}+1-g$, gives $2r\le \mathrm{deg}\mathcal{L}+2$, which is exactly the displayed Clifford inequality for curves.

The special case is impossible, and henceThis is the genus bound for a nonplanar degree-five curve. The geometric argument can be carried out after extension to an algebraic closure, which preserves these section dimensions.

For nonzero finite-dimensional spaces $V,W$ of sections of two line bundles on a smooth integral curve over an algebraically closed field, their product span in sections of the tensor product satisfies this bound. At one point choose bases with strictly increasing orders of vanishing. Multiplying every first-space basis vector by the first second-space vector, and then the last first-space vector by the remaining second-space vectors, produces distinct increasing orders and hence independent sections. This proves the dimension inequality needed for the Clifford inequality for curves.