I deleted a previous wrong and misleading answer.

The Moreau-Yoshida envelope is a special case of th infimal convolution of two convex functions $f$ and $g$ which is defined as
$$
f\Box g(x) = \inf_{y\in Y} f(y) + g(x-y).
$$
In other words: The infimal convolution of $f$ and $g$ is the largest (extended) real valued functional whose epigraph contains the (Minkowski) sum of the epigraphs of $f$ and $g$. Consequently, it is convex and lsc in $f$ and $g$ are. 

A more verbatim argument for the convexity: In general if $F:X\times X\to ]-\infty,\infty]$ is convex, then $f(x) = \inf_y F(x,y)$ is convex: Take $x_1$ and $x_2$ such that $f(x_i)$ is finite and $\xi_1> f(x_1)$, $\xi_2> f(x_2)$. Then there exist $y_1$, $y_2$ such that $F(x_i,y_i)<\xi_i$. By convexity of $F$ it holds for $0<\lambda<1$ that
$$\begin{array}{rl}
f(\lambda x_1 + (1-\lambda)x_2) & \leq F(\lambda x_1 + (1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2)\\
& \leq \lambda F(x_1,y_1) + (1-\lambda)F(x_2,y_2)\\
& \leq \lambda \xi_2 + (1-\lambda)\xi_2.
\end{array}
$$
Letting $\xi_i\to f(x_i)$ shows convexity of $f$. Now apply this to $F(x,y) = f(y) + g(x-y)$.