= Solution
The <orthogonal complement> $x^\perp$ is infinite-dimensional: otherwise $X=\operatorname{span}\{x\}\oplus x^\perp$ would be finite-dimensional. Inductively choose an <orthonormal sequence> $(e_n)$ in $x^\perp$, each new vector being orthogonal to the preceding finite <linear span>. The <Bessel inequality> gives $\sum_n|\langle e_n,z\rangle|^2\leq\|z\|^2$ for every $z\in X$, so $\langle e_n,z\rangle\to0$. Thus $(e_n)$ is a <weakly null orthonormal sequence>.
Set
$$
\boxed{x_n=x+\sqrt{1-\|x\|^2}\,e_n.}
$$
By <orthogonality>, $\|x_n\|^2=\|x\|^2+(1-\|x\|^2)=1$, while $\langle x_n-x,z\rangle\to0$ for every $z$. Hence $x_n\rightharpoonup x$. When $\|x\|=1$, this formula simply gives the constant sequence $x_n=x$. In particular, the <weak closure of the unit sphere> contains the whole <closed unit ball>.
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