Take any affine space $(A,\vec A)$, with the vector space $\vec A$ endowed by a topology $T_{\vec A}$ making $\vec A$ a topological vector space. Fix any $a_0\in A$ and let
$$T_A:=\{a_0+V\colon V\in T_{\vec A}\},
$$
where $a_0+V:=\{a_0+v\colon v\in V\}$.
Then $T_A$ is a topology over $A$ (because the map $\vec A\ni v\mapsto a_0+v\in A$ is a bijection). Moreover, the action
$$A\times\vec A\ni(a,v)\mapsto a+v\in A$$
of $\vec A$ on $A$ is continuous, as desired.

**Details on the latter sentence:** Take any $a_0+V\in T_A$, so that $V\in T_{\vec A}$.
Take then any $(a,v)\in A\times\vec A$ such that $a+v\in a_0+V$. By the transitivity of the action of $\vec A$ on $A$, there is some vector $v_a\in\vec A$ such that $a=a_0+v_a$. By the associativity of the action, $a_0+(v_a+v)=(a_0+v_a)+v=a+v\in a_0+V$, whence $$v_a+v\in V.$$
So, by the continuity of the addition in the topological vector space $\vec A$, there are members $V_1$ and $V_2$ of the topology $T_{\vec A}$ such that $0\in V_1$, $0\in V_2$, and
$$v_a+V_1+v+V_2\subseteq V.$$
Using now the associativity of the action again, we have
\begin{equation}
(a+V_1)+(v+V_2)=a_0+(v_a+V_1+v+V_2)\subseteq a_0+V. \tag{1}
\end{equation}
Since $0\in V_2$ and $0\in V_1$, we have $v\in v+V_2\in T_{\vec A}$ and $a\in a+V_1$. Also, $a+V_1=a_0+(v_a+V_1)$ and $v_a+V_1\in T_{\vec A}$, so that $a\in a+V_1\in T_A$. Thus, (1) shows that the action
$$A\times\vec A\ni(a,v)\mapsto a+v\in A$$
of $\vec A$ on $A$ is indeed continuous.