= Solution
For the centered ball, the <Hahn-Banach theorem> supplies a <norm>-one real functional with $f(y)=\|y\|>r$. Hence $f(z)\leq r$ on the ball, and any $t\in(r,\|y\|)$ separates it strictly from $y$.
For the general center, apply the same construction to $y-x$: choose $\|f\|=1$ and $f(y-x)=\|y-x\|$. Then
$$
f(z)\leq f(x)+r\quad(z\in B),\qquad
f(y)=f(x)+\|y-x\|>f(x)+r.
$$
For example,
$$
\boxed{t=f(x)+\frac{r+\|y-x\|}{2}}
$$
gives $f(z)<t<f(y)$ for every $z\in B$. The strict gap uses exactly the assumption that $y$ lies outside the closed ball.
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