Proof of The Induced Norm is a Norm, and Induces a Metric

Throughout, $\lVert w\rVert$ denotes the unique real number with $0\le\lVert w\rVert$ and $\lVert w\rVert^{2}=\langle w,w\rangle$, as in Norm Induced by a Complex Inner Product. We use two facts about real numbers, both established as in the proof of Properties of Complex Conjugation and Modulus: two nonnegative real numbers with equal squares are equal, by Existence and Uniqueness of the Nonnegative Square Root; and if $p,q$ are nonnegative reals with $p^{2}\le q^{2}$, then $p\le q$. We also use that a product of nonnegative reals is nonnegative, by the second order-compatibility condition of ordered fields.

Claim 1. By Cauchy-Schwarz Inequality in a Complex Inner Product Space and the definition of the induced norm,

Both $|\langle u,v\rangle|$ and $\lVert u\rVert\lVert v\rVert$ are nonnegative, so $|\langle u,v\rangle|\le\lVert u\rVert\,\lVert v\rVert$.

Claim 2. Positivity. $0\le\lVert v\rVert$ holds by definition; and if $\lVert v\rVert=0$, then $\langle v,v\rangle=\lVert v\rVert^{2}=0$, so $v=0_{V}$ by claim 4 of Elementary Properties of a Complex Inner Product.

Absolute homogeneity. Let $\lambda$ be a complex number. By condition 3 of Complex Inner Product Space, claim 2 of Elementary Properties of a Complex Inner Product, and $\overline{\lambda}\lambda=|\lambda|^{2}$ (claim 3 of Properties of Complex Conjugation and Modulus),

Both $\lVert\lambda v\rVert$ and $|\lambda|\lVert v\rVert$ are nonnegative, so $\lVert\lambda v\rVert=|\lambda|\,\lVert v\rVert$.

Triangle inequality. By additivity in each argument (condition 2 of Complex Inner Product Space and claim 1 of Elementary Properties of a Complex Inner Product),

By condition 1 of Complex Inner Product Space, $\langle v,u\rangle=\overline{\langle u,v\rangle}$, so claim 2 of Properties of Complex Conjugation and Modulus gives $\langle u,v\rangle+\langle v,u\rangle=2\operatorname{Re}\langle u,v\rangle$, with the real part. By claim 6 of Properties of Complex Conjugation and Modulus and claim 1 above,

Adding this inequality to itself and then adding $\lVert u\rVert^{2}+\lVert v\rVert^{2}$, both steps permitted by the first order-compatibility condition,

Both $\lVert u+v\rVert$ and $\lVert u\rVert+\lVert v\rVert$ are nonnegative, so $\lVert u+v\rVert\le\lVert u\rVert+\lVert v\rVert$. Thus $\lVert\cdot\rVert$ satisfies the three conditions of Norm on a Complex Vector Space.

Claim 3. We check the four conditions of Metric Space for $d(u,v)=\lVert u-v\rVert$, a real number for all $u,v\in V$. First, $0\le\lVert u-v\rVert$ by claim 2. Second, if $\lVert u-v\rVert=0$ then $u-v=0_{V}$ by claim 2, and adding $v$ to both sides gives $u=v$ by conditions 1, 3 and 4 of Vector Space over a Field and claim 2 of Elementary Identities in a Vector Space; conversely $\lVert u-u\rVert=\lVert 0_{V}\rVert=0$, since $\langle 0_{V},0_{V}\rangle=0$ by claim 3 of Elementary Properties of a Complex Inner Product. Third, by conditions 7 and 5 of Vector Space over a Field and claim 5 of Elementary Identities in a Vector Space,

so by absolute homogeneity and claim 8 of Properties of Complex Conjugation and Modulus, which gives $|-1|=1$,

Fourth, for $u,v,w\in V$ we have $u-w=(u-v)+(v-w)$, by conditions 1, 2, 3 and claim 2 of Elementary Identities in a Vector Space, so the triangle inequality of claim 2 gives $\lVert u-w\rVert\le\lVert u-v\rVert+\lVert v-w\rVert$. Hence $d$ is a metric on $V$.