It is relatively easy to prove the convexity of $$\inf_a\{ [x+\sin(x+a)]^2\}, a \in [-\pi, \pi]$$ by finding the zero of $\partial_a[x+\sin(x+a)]^2$: $$\frac{\partial}{\partial a}[x+\sin(x+a)]^2 = 2cos(x+a)[x+\sin(x+a)] = 0\Rightarrow a=\pm\frac{\pi}{2}-x \text{ or } 2n\pi-x-asin(x)$$ and then substituting back to remove $\inf_a$.
This is harder for the 2d extension: $$\inf_a\{ [x+\sin(x+a)]^2 + [y+\sin(y+a)]^2\}, a \in [-\pi, \pi]$$
because the $a(x,y)$ is no longer easy to solve for. Graphing shows that the function is non-convex.
Is it possible to remove the $\inf_a$ or prove the convexity for the second function without finding zeros for $$ \frac{\partial}{\partial a}\left\{[x+\sin(x+a)]^2 + [y+\sin(y+a)]^2\right\}\text{?}$$ I'm not sure how to deal with the infimum, or the non-linearity in both $x$ and $a$.
A simplification of the 1d case which could help in the 2d case. (Since your graph of the latter already shows that it is not convex I am not quite sure what you are trying to achieve.)
Since for each $x\in\mathbb R$ the function $[-\pi,\pi]\ni a\mapsto\sin(x+a)\in [-1,1]$ is surjective we can simplify the infimum to $$\tag{1} f(x):=\inf_{a\in\mathbb R}\big[x+\sin(x+a)\big]^2=\inf_{b\in[-1,1]}[x+b]^2\,. $$ Splitting up $x$ into cases shows that the infimum over $b$ is attained by the following $b\,:$ \begin{align} x&\in [-1,1]: & b=-x\,,\\ x&\in (1,\infty): & b=-1\,,\\ x&\in (-\infty,-1): & b=+1\,. \end{align} Therefore, $f$ is the convex function $$ f(x)=\begin{cases}(x+1)^2\,,&x\in(-\infty,-1)\,,\\[2mm]0\,,&x\in[-1,1]\,,\\[2mm](x-1)^2\,,&x\in(1,\infty)\,.\end{cases} $$