= Solution
Assume \b[$\operatorname{argmin}f\ne\varnothing$], as is necessary for convergence to a minimizer. The printed hypotheses alone omit this condition: $f(x,y)=x$ in scalar $x$ is convex with an $L$-<Lipschitz gradient> for any prescribed $L>0$, but has no minimizer, and the update sends $x^k=x^0-k\tau$ to $-\infty$.
Under existence, define the full update $T(x,y)=(x^+,y^+)$. The proximal $y$-step is uniquely defined by strong convexity of its quadratic term and satisfies
$$
y-y^+=\tau\nabla_y f(x,y^+),\qquad x-x^+=\tau\nabla_x f(x,y^+).
$$
The <partial subdifferential> in this equation belongs to the scalar function $y\mapsto f(x,y)$; differentiating a subdifferential itself would not define the printed proximal minimization.
We prove directly that the <alternating proximal-gradient operator> $T$ is averaged. For two inputs $z=(x,y)$ and $\widetilde z=(\widetilde x,\widetilde y)$, set
$$
w=(x,y^+),\quad\widetilde w=(\widetilde x,\widetilde y^+),\quad
p=x-\widetilde x,\quad q=y^+-\widetilde y^+,\quad
(a,b)=\nabla f(w)-\nabla f(\widetilde w).
$$
The input difference is $(p,q+\tau b)$, the output difference is $(p-\tau a,q)$, and the difference of the two update residuals is $\tau(a,b)$. Exact expansion gives
$$
\begin{aligned}
\|Tz-T\widetilde z\|_2^2
&=\|z-\widetilde z\|_2^2-2\tau[\langle p,a\rangle+\langle q,b\rangle]
+\tau^2(\|a\|_2^2-\|b\|_2^2)\\
&\leq\|z-\widetilde z\|_2^2
-\left(\frac{2\tau}L-\tau^2\right)\|a\|_2^2
-\left(\frac{2\tau}L+\tau^2\right)\|b\|_2^2\\
&\leq\|z-\widetilde z\|_2^2
-\left(\frac2{\tau L}-1\right)\|(I-T)z-(I-T)\widetilde z\|_2^2.
\end{aligned}
$$
The first inequality uses <cocoercivity> at the mixed points $w,\widetilde w$. Therefore \b[$T$ is $(\tau L/2)$-averaged] for the requested $0<\tau L\leq1$. In particular, the endpoint $\tau L=1$ is covered: $T$ is <firmly nonexpansive>, not merely nonexpansive.
A <fixed point> has $y^+=y$ and $x^+=x$, so the two optimality equations give $\nabla_y f=0$ and $\nabla_x f=0$. Conversely, at a minimizer both gradients vanish and neither step moves. By <convexity>,
$$
\boxed{\operatorname{Fix}T=\operatorname{argmin}f}.
$$
The assumed existence of a minimizer supplies a <fixed point>. The <Browder convergence theorem for averaged operators> now gives
$$
\boxed{(x^k,y^k)\longrightarrow(x^*,y^*)\in\operatorname{argmin}f}.
$$
No uniqueness of the minimizer is needed. The same estimate even proves averagedness for $0<\tau L<2$ in this smooth setting, but the requested interval already suffices. The existence condition and the use of the gradient at $(x^k,y^{k+1})$ are both essential to the proof as presented.
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