= Solution
Fix $i$ and let $M_i=\overline{\operatorname{span}}\{x_j:j\ne i\}$. The hypothesis gives $M_i\ne X$. Moreover $x_i\notin M_i$, since otherwise the closed span of all the <vectors> would be contained in $M_i$. Hence $d_i=\operatorname{dist}(x_i,M_i)>0$.
On $M_i+\mathbb Rx_i$, define $\phi_i(m+tx_i)=t$. It is well-defined because $x_i\notin M_i$. For $t\ne0$,
$$
\|m+tx_i\|=|t|\|x_i+m/t\|\geq|t|d_i,
$$
so $|\phi_i(m+tx_i)|\leq\|m+tx_i\|/d_i$; the inequality also holds for $t=0$. The <Hahn-Banach theorem> extends $\phi_i$ without increasing its <norm> to $x_i^*\in X^*$. This gives $x_i^*(x_j)=\delta_{ij}$.
If two continuous functionals have these values, their difference vanishes on the <linear span> of every $x_j$, which is dense in $X$. <Continuity> then makes the difference zero everywhere. Thus each functional is unique, proving the required \b[unique biorthogonal system]. In fact its <norm> is $\|x_i^*\|=1/d_i$, by taking <vectors> in $M_i$ approaching the distance infimum. This is the coordinate-functional construction for a <minimal total family in a normed space>.
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