= Solution
The continuous <dual space> $X^*$ consists of all bounded real <linear functionals> on $X$, with <operator norm> $\|f\|=\sup_{\|x\|\le1}|f(x)|$. Let $(f_n)$ be Cauchy in this <norm>. For each $x$, $|f_n(x)-f_m(x)|\le\|f_n-f_m\|\|x\|$, so $f_n(x)$ converges in $\mathbb R$; define its limit to be $f(x)$.
Pointwise limits of the linear identities show that $f$ is linear. The <sequence> of <norms> is bounded, say by $M$, so $|f(x)|\le M\|x\|$ and $f\in X^*$. Given $\varepsilon>0$, choose $N$ so that $\|f_n-f_m\|\le\varepsilon$ for all $m,n\ge N$. Holding $n$ fixed and taking the pointwise limit in $m$ gives $|f_n(x)-f(x)|\le\varepsilon\|x\|$ for all $x$. Thus $\|f_n-f\|\le\varepsilon$, proving
$$
\boxed{X^*\text{ is a Banach space}}.
$$
Completeness of $X$ is not needed; completeness of the scalar <field> supplies the pointwise limit.
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