Characterization of compact subsets.
If $K\subseteq X$, and $x\in \overline{K\cap \mathbb R}\backslash (K\cap \mathbb R)$ (with the closure taken relative to $\mathbb R$), then let $k_n\to x$ be a sequence with $k_n\in K\cap \mathbb R$. Then if $U_n=X\backslash \left(\{k_m\mid m\geq n\}\cup \{x\}\right)$, then $\{U_n\}$ is an open cover of $K$ with no finite subcover, hence $K$ is not compact.
Similarly, if $K\cap \mathbb R$ is unbounded, then letting $k_n\in K$ with $k_n$ converging to $\pm \infty$ (in the usual sense, not to the point $\infty\in X$), we see that setting $U_n=X\backslash \{k_m\mid m\geq n\}$ again gives an open cover $\{U_n\}$ of $K$ with no finite subcover, so $K$ is again not compact.
Thus every compact set intersects $\mathbb R$ in a closed and bounded, i.e., Euclidean compact, set. Conversely, if $K\cap \mathbb R$ is Euclidean compact, then certainly $K$ is compact, as at most one point is added to $K\cap \mathbb R$ to produce $K$, so the compact subsets of $X$ are precisely those of the form $K_e$ or $K_e\cup\{\infty\}$ for some Euclidean compact $K_e$.
Generation by countably many compact sets.
From the above, we immediately see that the sequence $K_n:=[-n,n]\cup\{\infty\}$ is an increasing sequence of compact sets such that each compact $K\subset X$ must lie in some $K_n$. Moreover, we claim the sets $K_n$ generate the topology on $X$. To see this, suppose $A\subset X$, with $A\cap K_n$ open for each $n$. If $\infty\in A$, then $A\cap [-n,n]$ is cocountable and open (relative to $[-n,n]$) for each $n$, whereby $A\cap \mathbb R$ is cocountable and Euclidean open, hence $A$ is open. If $\infty\notin A$, then $A\cap [-n,n]$ is still Euclidean open for each $n$, whereby $A$ is open in $\mathbb R$.
$X$ is not paracompact.
Let $U_n=(n-\frac{1}{3},n+\frac{1}{3})$ for $n\in \mathbb N$, $U_\infty=X\backslash \mathbb N$. Then $\{U_n\}\cup\{U_\infty\}$ is an open cover of $X$ with no locally finite refinement, since any such refinement would necessarily include one open subset $V_n\subseteq U_n$, in order to include each $n$, but then every neighborhood of $\infty$ intersects each $V_n$. (Since this cover was countable, we have shown further that $X$ is not even countably paracompact.)
$X$ is metacompact.
Let $\mathcal U$ be an open cover of $X$, then by meta-compactness of $\mathbb R$ the family $\mathcal V:=\{U\cap \mathbb R\mid U\in\mathcal U\}$ has an open refinement $\mathcal V'$ covering $\mathbb R$ that is point-finite. Adding a single member $U\in\mathcal U$ containing $\infty$ to $\mathcal V'$ gives a point-finite refinement $\{U\}\cup \mathcal V'$ of $\mathcal U$.
$X$ is not locally compact.
In fact, from the characterization of compact subsets above, $X$ is not even weakly locally compact, i.e., $\infty$ has no compact neighborhood, since every neighborhood of $\infty$ intersects $\mathbb R$ in an unbounded set, which is therefore not Euclidean compact.
Remark on separation properties.
Though not in the category of compactness properties per se, it is worth noting that since $X$ is clearly $T_1$, understanding the compact subsets of $X$ allows us to apply this result to further understand the separation properties of $X$.
That is, from the description of compact sets, we easily have that any family of compact subsets of $X$ with the finite intersection property has nonempty intersection. That is, $X$ has the "compact intersection property" ($KIP$) defined in the linked question, and so that answer implies that $X$ is $k_2$-Hausdorff.
On the other hand, $X$ is not weak Hausdorff, since compact subsets of $\mathbb R$ are images of compact Hausdorff spaces, yet are generally not closed in $X$, as they may be uncountable.
