Using the standard topology of $\mathbb{R}$ is sufficient to ensure that a compact subset of $\mathbb{R}$ is closed and bounded and vice versa. A standard topology of $\mathbb{R}$ is Hausdorff. I wonder if there is a nonHausdorff topology on $\mathbb{R}$ such that with respect to which a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded.
Note: Thanks to the contributors' works. I made a mistake. What I am after is a nonHausdorff topology $\tau$ on $\mathbb{R}$ such that, if $\tau'$ is the usual topology on $\mathbb{R}$, a subset of $\mathbb{R}$ is $\tau$-compact if and only if it is $\tau$-closed and $\tau'$-bounded. It is great to know the existence of that type of nonHausdorff topology ($\tau$-bounded) as given in an answer below too.
Consider the equivalence relation generated by $x\sim y$ if they both are in $[0,1]$. Then let $\tau$ be the topology saturated by such relation (i.e. take the quotient space and then pull back the quotient topology). Explicitly, you declare a set open if it is open in Euclidean topology and if, whenever it contains a point in $[0,1]$ then it contains the whole $[0,1]$.
Such topology in not hausdorff because points in $[0, 1]$ are indistinguishable. But from the fact that the quotient space is homeo to $\mathbb R$ you get the desired property.