= Solution
Write $\psi=\psi_{x_0}$; its value at $x_0$, if that point is in $(0,1)$, may be assigned arbitrarily because <Lp spaces> identify <functions> equal <almost everywhere>. Its <L-infinity norm> is one.
The supplied estimate shows that $\psi$ cannot agree <almost everywhere> with any <continuous function> $q$: choosing $g=-q$ would give $0\geq|q(x_0)|+1$. Hence
$$
V=C([0,1])\oplus\langle\psi\rangle
$$
is a <direct sum>, so every element has a unique expression $g+c\psi$.
For any complex $a$ with $|a|\leq1$, define the <linear functional>
$$
L_a(g+c\psi)=g(x_0)+ca.
$$
If $c\ne0$, the supplied estimate, applied to $g/c$, and <absolute homogeneity of a norm> give
$$
\|g+c\psi\|_\infty=|c|\|g/c+\psi\|_\infty
\geq|g(x_0)|+|c|
\geq|g(x_0)+ca|.
$$
For $c=0$, the same bound follows from <point evaluation>. Thus $\|L_a\|\leq1$, and $L_a(1)=1$ makes its <operator norm> exactly one. The <complex Hahn-Banach theorem> extends $L_a$ to a norm-one <linear functional> on all of $X$, with value $a$ at $\psi$.
Conversely every norm-one extension satisfies $|\ell_{x_0}(\psi)|\leq\|\psi\|_\infty=1$. We have therefore determined all possible values:
$$
\boxed{\{\ell_{x_0}(\psi_{x_0}):\ell_{x_0}\text{ is a norm-one extension of }E_{x_0}\}
=\{a\in\mathbb C:|a|\leq1\}.}
$$
In particular, taking $a=0$ and $a=1$ proves <nonuniqueness of norm-preserving point evaluation extensions>.
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