= Solution
Take $Y=X^{**}$, which is a <Banach space> by the completeness of the dual just proved. The <canonical embedding into the bidual> is
$$
\Phi(x)(f)=f(x),\qquad f\in X^*.
$$
For fixed $x$ this is a bounded <linear functional> on $X^*$, since $|f(x)|\le\|f\|\|x\|$, and $\Phi$ is linear in $x$. This gives $\|\Phi(x)\|\le\|x\|$. For $x\ne0$, the norming functional from the <Hahn-Banach theorem> has $\|f\|=1$ and $f(x)=\|x\|$, giving the reverse inequality. At zero equality is immediate. Thus
$$
\boxed{\|\Phi(x)\|=\|x\|\text{ for all }x\in X}.
$$
In particular $\Phi$ is injective and isometric. If a dense isometric embedding is wanted, rather than merely the stated embedding, take the <norm> closure of $\Phi(X)$ in $X^{**}$; a closed subspace of a <Banach space> is complete and supplies a <completion of a normed space>.
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