= Solution
The <continuous dual space> $X^*$ consists of bounded linear functionals with <operator norm> $\|v\|=\sup_{\|x\|\le1}|v(x)|$. For an operator-norm <Cauchy sequence> $(v_n)$, each $(v_n(x))$ is Cauchy in $\mathbb R$. Define $v(x)=\lim_n v_n(x)$. It is linear, and bounded because the Cauchy sequence has uniformly bounded norms. Passing to the pointwise limit in $|(v_n-v_m)(x)|\le\varepsilon\|x\|$ gives $\|v_n-v\|\le\varepsilon$. Thus $X^*$ is a <Banach space>, even when $X$ is incomplete.
The canonical map $\phi(x)(v)=v(x)$ is linear and satisfies $\|\phi(x)\|\le\|x\|$. For $x\ne0$, the norm-one functional $tx\mapsto t\|x\|$ on its span extends by the <Hahn-Banach theorem> to a norm-one functional on $X$. Hence $\|\phi(x)\|\ge\|x\|$, proving that $\phi$ is an <isometry> and therefore <injective>.
For a nonsurjective example take $X=c_0$ with the supremum norm. It is complete as a closed subspace of $\ell^\infty$. Every functional is given by an $\ell^1$ sequence: finite-coordinate sign tests give summability of its coefficients, and truncation of each $c_0$ vector gives the series representation. Conversely any $\ell^1$ sequence defines such a bounded functional. Similarly $(\ell^1)^*=\ell^\infty$, so $c_0^{**}=\ell^\infty$. The canonical image is the null sequences; the constant sequence one is outside it. Thus \b[$c_0$ is a nonreflexive Banach space].
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