= Solution
For a step size $\tau>0$, define the set-valued maps
$$
\boxed{F_{\tau f}=I-\tau\partial f,\qquad B_{\tau f}=(I+\tau\partial f)^{-1}.}
$$
The <forward subgradient step> maps $x$ to the set $\{x-\tau p:p\in\partial f(x)\}$. If $f$ is differentiable this is the explicit gradient step $x-\tau\nabla f(x)$. The <backward subgradient step> consists of the $y$ satisfying $x-y\in\tau\partial f(y)$, an implicit step for the <subgradient> flow. It is the <resolvent of a monotone operator> associated with $\partial f$.
Suppose $y_1,y_2\in B_{\tau f}(x)$. Then $p_i=(x-y_i)/\tau\in\partial f(y_i)$. The two defining <subgradient inequalities> are
$$
f(y_2)\geq f(y_1)+\langle p_1,y_2-y_1\rangle,\qquad
f(y_1)\geq f(y_2)+\langle p_2,y_1-y_2\rangle.
$$
Their sum proves <monotonicity of a convex subdifferential>, $\langle p_1-p_2,y_1-y_2\rangle\geq0$. But $p_1-p_2=-(y_1-y_2)/\tau$, so
$$
0\leq-\frac1\tau\|y_1-y_2\|^2,\qquad\boxed{y_1=y_2}.
$$
Convexity supplies the <subgradient> inequalities and monotonicity; membership in the <subdifferential> ensures the two function values are finite, so subtraction is legitimate. Positivity of $\tau$ supplies the decisive sign. Properness rules out the identically infinite and negative-infinity pathologies in the overall setting, but <lower semicontinuity> is not needed for this at-most-one argument. Its role is in existence, proved next. The backward step cannot have two values, though uniqueness alone has not yet shown its domain is all of $\mathbb R^n$.
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