Is the action of a finite group always discontinuous? Let $G$ be a finite group acting on a topological space $X$. Is it true that the action of $G$ is always proper?
To say that it is proper we have to show that the map $\theta : G \times X \longrightarrow X \times X$ given by $(g,x) \mapsto (x,gx)$ is a proper map (inverse image of a compact set is compact)
I can show this is true when $X$ is a compact Hausdorff space but I am unsure how to prove it in general. 
To find a counterexample I thought along the lines that if the action is proper then since $G$ is discrete the action must be properly discontinuous. So I should find an example of a finite group action that is not discontinuous and I will be done. But I can't come up with anything. It makes me wonder if it is true that a finite group action on topological space is always discontinuous?
An action is discontinuous if for each $x\in X$ we can find a neighbourhood $U$ of $x$ such that $gU\cap U$ is empty unless $g$ fixes $x$.
 A: The action of a finite group $G$ on a Hausdorff space $X$ is always discontinuous.
Let $x_0 \in X$, and consider the orbit $Gx_0 = \{ x_0, x_1,\dotsc, x_k\}$. Since $X$ is Hausdorff, we can find open neighbourhoods $V_{\kappa}$ of $x_{\kappa}$ such that $V_{\kappa} \cap V_{\lambda} = \varnothing$ for $\kappa \neq \lambda$. For each $g \in G$, we can find an open neighbourhood $W_g$ of $x_0$ such that $gW_g \subset V_{\kappa}$ if $gx_0 = x_{\kappa}$ since $x \mapsto gx$ is continuous. (For example $W_g = V_0 \cap g^{-1}V_{\kappa}$.) Then
$$U = V_0 \cap \bigcap_{g \in G} W_g$$
is an open neighbourhood of $x_0$, and we have $gU \subset V_{\kappa}$ whenever $gx_0 = x_{\kappa}$, in particular
$$gU \cap U \neq \varnothing \iff gx_0 = x_0.$$
On a non-Hausdorff space, the action of a finite group need not be discontinuous. As an extreme example, take an indiscrete space, then the only discontinuous action is the trivial action.
If $X$ is Hausdorff, an action of a finite group is also proper: Given a compact $K\subset X\times X$, let $K_i = \pi_i(K) \subset X$. Then $K_1,K_2$ are compact, $K \subset K_1 \times K_2$, and $\theta^{-1}(K)$ is a closed subset of $\theta^{-1}(K_1 \times K_2)$. But
$$\theta^{-1}(K_1\times K_2) = \bigcup_{g\in G} \{g\} \times (K_1 \cap g^{-1}K_2)$$
is the union of finitely many compact sets, hence compact.
