Define a family of subsets of $X=\Bbb R^n$ by the following.
$F \in {\cal F} \iff F$ is $\emptyset$ or $\Bbb R^n$ or a bounded closed subset of with respect to the Euclidean topology.
(1) Show that ${\cal F}$ satisfies the axioms of closed sets. (So ${\cal F}$ defines a topology.)
I could have checked all of the axioms except the one that ${\cal F}$ is closed under finite union. We need to show that $$V:=\bigcup_{i=1}^n V_i\in {\cal F} \quad\forall V_i\in {\cal F}.$$ We can assume that $V_i \ne \emptyset$ for all $i$. If there exists $V_k$ so that $V_k=X$ then $V=X\in {\cal F}$. For otherwise, $$X-V = X-\left(\bigcup_{i=1}^n V_i\right)=\bigcap_{i=1}^n\left(X-V_i\right)$$ is open since $X-V_i$ is open for all $i$. What's left is to prove $V$ is bounded. I want to find a ball $B_r(x) \subset X$ so that $V \subset B_r(x)$. I can imagine its existence since $V_i \subset B_{r_i}(x_i)$ for some $r_i>0, x_i\in X$ but cannot find any appropriate candidate.
Let $T$ denote the topological space $(\Bbb R^n, {\cal F})$.
(2) Show that $T$ is not a Hausdorff space.
Suppose to the contrary that $T$ is a Hausdorff space. Then for all $x,y\in T$ we can find $U, V \in {\cal F}$ such that $x\in U, y\in V$ and $U \cap V =\emptyset$. I don't whether I'm on the right path but there's no contradiction comes out yet.
(3) Show that $T$ is connected.
I was completely lost at this point.
Any help would be much appreciated.
HINT: A set $A\subseteq\Bbb R^n$ is bounded if and only if there is an $r>0$ such that $A\subseteq B(\mathbf{0},r)$, where $\mathbf{0}$ is the origin. This makes it easy to show that a finite union of bounded sets is bounded.
For the other two parts, note that if $F\in\mathscr{F}$ and $F\ne\Bbb R^n$, then $F$ is compact in the usual topology on $\Bbb R^n$. (Why?) Use this observation to show that if $U$ and $V$ are non-empty open sets in this new topology on $\Bbb R^n$, then $U\cap V\ne\varnothing$; both $(2)$ and $(3)$ then follow easily.