= Solution
Write $h=f\mathbin\square g$ for the <infimal convolution>. It is proper by the permitted hypothesis. If $h(x_0)$ and $h(x_1)$ are finite, choose $y_0,y_1$ within $\varepsilon$ of their respective infima. For $0<t<1$, put $x_t=(1-t)x_0+tx_1$ and $y_t=(1-t)y_0+ty_1$. <Convexity> of $f$ and $g$ gives
$$
\begin{aligned}
h(x_t)&\leq f(x_t-y_t)+g(y_t)\\
&\leq(1-t)[f(x_0-y_0)+g(y_0)]+t[f(x_1-y_1)+g(y_1)]\\
&\leq(1-t)h(x_0)+th(x_1)+\varepsilon.
\end{aligned}
$$
Let $\varepsilon\downarrow0$. If either endpoint value is infinite, the desired inequality is automatic. Thus \b[the <infimal convolution> is convex], without assuming the infimum is attained.
For its <convex conjugate>, replace a negative infimum by a supremum and then change variables $u=x-y$:
$$
\begin{aligned}
h^*(p)
&=\sup_x\left[\langle p,x\rangle-\inf_y\{f(x-y)+g(y)\}\right]\\
&=\sup_{x,y}\{\langle p,x\rangle-f(x-y)-g(y)\}\\
&=\sup_{u,y}\{\langle p,u\rangle-f(u)+\langle p,y\rangle-g(y)\}\\
&=f^*(p)+g^*(p).
\end{aligned}
$$
The two suprema separate because $u$ and $y$ are independent. Properness of $f$ and $g$ makes each supremum strictly greater than $-\infty$, so this separation remains valid when one or both are $+\infty$. Therefore \b[$(f\mathbin\square g)^*=f^*+g^*$]. This is the <conjugate of an infimal convolution>. The bounded-domain assumption is unnecessary for these two calculations; the assumed properness and <lower semicontinuity> of $h$ will be used when applying <subgradient inversion under convex conjugacy>.
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