Let $F = \cup^{\infty}_{k=1} A_i$ where each $A_i$ is a closed set. Since $F$ is $F_\sigma$, every $F_\sigma$ set is measurable, and every measurable set is translation invariant, F is translation invariant. Because of this, $$F + y = \cup^{\infty}_{k=1} A_i+y$$ and hence the translation of $F$ is $F_\sigma$.
Here is my second try.
Let $A$ be a closed subset of $\mathbb{R}$. Since it is closed $\mathbb{R} \setminus A$ is open. That implies that there exists a countable collection of disjoint open intervals $\{I_k\}^{\infty}_{k=1}$ where $\mathbb{R} \setminus A = \cup^{\infty}_{k=1}I_k$. Now let us take the translation of $A$ with respect to $y$ as $$ A + y = \{ a + y : a \in A \}$$ It is closed if and only if the translation of every $I_k$ in $\{I_k\}^{\infty}_{k=1}$ is open with respect to the translation $y$.
Since every $F_\sigma$ set is the union of a countable set of closed sets, and its translation is the translation of each closed set, the translation of every $F_\sigma$ set in $\mathbb{R}$ is an $F_\sigma$ set.
