Let $\mathbb{R^n}$ be a topological space where we define topology $\tau$ as follows: a subset $A \subset \mathbb{R^n}$ is closed if it is compact in the standard (Euclidean) topology, or if $A = \mathbb{R^n}$ (This means that open subsets in $\tau$ are those whose complement is compact in standard topology.)
Show that $F \subset \mathbb{R^n}$ is compact in $\tau $ $\Leftrightarrow $ $F$ is closed in standard topology.
My proof:
($\Rightarrow$)
Let $F$ be compact in $\tau$. Let $U $ be a open (in $\tau$) cover of $F$. Because $F$ is compact there are open sets $U_1, U_2, \ldots , U_n$ so that $F = U_1 \cup U_2 \cup \ldots \cup U_n $.
For each $U_i$ is $U_i^c$ compact in standard topology. A union of a finite number of compact sets is compact. So $F$ is compact in standard topology. Because $\mathbb{R}$ is a metric space, a compact set $F$ is closed. (In standard topology.)
($\Leftarrow$)
Let $F$ be closed in standard topology.
Here I am a bit lost as for how to continue. We know that $F^c$ is open (in s.t.)
Let $U$ be an open cover for $F$ in $\tau$. Can we find a finite number of open sets that cover $F$? Those need to have compact complements in standard topology. Can we find a finite number of compact sets that make up a whole $F$?
There was a similar question asked a while ago, but I did not completely understand the answer.
As SvanN says in the comments, your "topology" doesn't form a topology.
The open subsets of $\Bbb{R}^n$ are not closed under arbitrary intersections: $$\bigcap_{n=1}^\infty \left(\frac{-1}{n},\frac{1}{n}\right) = \{0\},$$ so the open subsets of $\Bbb{R}^n$ cannot be the closed sets of a topology.
Instead, the question you've linked suggests you should be defining your topology as $U$ is open in $\tau$ if $U^C$ is compact in the usual topology, plus the null set of course.
Then the question you need to answer is why this definition of $\tau$ gives a topology.
After that, you can try to reread the answer to the linked question, and it should hopefully make more sense.