The induced length is a norm

Nonnegativity and separation are [A1]: $\mathrm{\parallel }v\mathrm{\parallel }\ge 0$, and $\mathrm{\parallel }v\mathrm{\parallel }=0$ exactly when $⟨v,v⟩=0$, that is exactly when $v=0$.

Homogeneity: $⟨\lambda v,\lambda v⟩=\lambda \bar{\lambda }⟨v,v⟩=\mathrm{\mid }\lambda {\mathrm{\mid }}^2\mathrm{\parallel }v{\mathrm{\parallel }}^2$ by sesquilinearity and [A4], and both $\mathrm{\parallel }\lambda v\mathrm{\parallel }$ and $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$ are nonnegative with equal squares, so $\mathrm{\parallel }\lambda v\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }v\mathrm{\parallel }$ by uniqueness of the nonnegative square root.

Triangle inequality: expanding and using conjugate symmetry gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2+2\mathrm{Re}⟨u,v⟩+\mathrm{\parallel }v{\mathrm{\parallel }}^2$, and $\mathrm{Re}z\le \mathrm{\mid }z\mathrm{\mid }$ for every scalar $z$, since either $\mathrm{Re}z<0\le \mathrm{\mid }z\mathrm{\mid }$ or $\mathrm{Re}z=a\ge 0$ with $a^2\le a^2+b^2=\mathrm{\mid }z{\mathrm{\mid }}^2$ by [A6]; with [A2] this gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2\le \mathrm{\parallel }u{\mathrm{\parallel }}^2+2\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }+\mathrm{\parallel }v{\mathrm{\parallel }}^2=(\mathrm{\parallel }u\mathrm{\parallel }+\mathrm{\parallel }v\mathrm{\parallel }{)}^2$.

Both sides of the inequality in step 1.3 are nonnegative, so monotonicity of squaring on the nonnegatives turns it into $\mathrm{\parallel }u+v\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }+\mathrm{\parallel }v\mathrm{\parallel }$.

Steps 1.1, 1.2 and 2.1 are exactly the clauses of [A3] over $\mathbb{R}$, and over $\mathbb{C}$ they are the same clauses with the complex modulus, so $\mathrm{\parallel }\cdot \mathrm{\parallel }$ is a norm on $V$ in either scalar field.