= Solution
Fix $x$ and minimize $J_x(y)=f(y)+\|y-x\|^2/(2\tau)$. A proper lower semicontinuous <convex function> has an <affine minorant> $f(y)\geq a\cdot y+b$, as follows by separating a point below its closed <epigraph>. Hence
$$
J_x(y)\geq a\cdot y+b+\frac1{2\tau}\|y-x\|^2\longrightarrow+\infty
\quad\text{as }\|y\|\to\infty.
$$
The quadratic dominates the linear term, proving <coercivity>. Properness supplies at least one finite trial value, <lower semicontinuity> passes to limits, and finite-dimensional compactness makes a bounded minimizing sequence converge along a subsequence to a minimizer. Thus the <proximal operator> exists at every $x$.
The <subdifferential sum rule> applies because the quadratic is finite and continuous everywhere. The <Fermat rule for convex minimization> gives
$$
0\in\partial J_x(y)=\partial f(y)+\frac{y-x}{\tau}
\quad\Longleftrightarrow\quad x\in y+\tau\partial f(y).
$$
Thus the minimizer lies in $B_{\tau f}(x)$. The uniqueness proved in part (a), or strict <convexity> of the quadratic sum, now yields
$$
\boxed{B_{\tau f}(x)=\left\{\operatorname{prox}_{\tau f}(x)\right\}\quad
\text{for every }x\in\mathbb R^n.}
$$
This proof displays the separate roles of properness, <lower semicontinuity>, <convexity> and finite dimension. In particular, compactness here is not inferred merely from strict <convexity>.
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