Compactness properties of rational sequence topology For each irrational ${x}$ we take a sequence of rational numbers ${x_k}$ with the property that ${x_k}$ converges to it in the Euclidean topology.
The rational sequence topology is given by defining each rational number singleton to be open, and using ${\displaystyle U_{n}(x):=\{x_{k}:n\leq k<\infty \}\cup \{x\}}$ as a basis for irrational point ${x}$.

*

*Proof that this topology is not metacompact but countably metacompact.


*Will it be countably paracompact ?
(this example is example #65 in Steen, L. A.; Seebach, J. A. (1995), Counterexamples in Topology )
 A: $X$ is not metacompact/paracompact/metaLindelöf
To see that $X$ is not metacompact: consider the open cover $\mathcal{U} = \{\{x\}: x \in \mathbb{Q}\} \cup \{U_0(x): x \notin \mathbb{Q}\}$.
Note that all members of $\mathcal{U}$ are countable ($X$ is a so-called locally countable space).
Also note that $\mathbb{Q}$ is a countable dense set.
If $\mathcal{U}$ would have a point-finite open refinement $\mathcal{V}$ that refines $\mathcal{U}$ then define  
$$\mathcal{V}_x = \{O \in \mathcal{V}: x \in O\}$$
and by definition all $\mathcal{V}_x$ are finite sets (it is enough for them to be countable, in fact).
Now because every member of the open cover $\mathcal{V}$ contains a rational point:
$$\mathcal{V} = \bigcup_{q \in \mathbb{Q}} \mathcal{V}_q$$ 
which is a countable union of finite sets, so $\mathcal{V}$ is a countable cover. For every $V \in \mathcal{V}$ we can pick $U(V) \in \mathcal{U}$ with $V \subseteq U(V)$, by the definition of refinement, and then 
$$\{ U(V): V \in \mathcal{V}\} \subset \mathcal{U}$$
is a countable subcover of $\mathcal{U}$. But all members of $\mathcal{U}$ are countable and this would thus imply $\mathbb{R}$ is countable, contradiction.
So $\mathcal{U}$ does not have a point-countable, point-finite, or locally finite refinement (each property implies the previous one), so $X$ is neither metalindelöf, metacompact, or paracompact (already non-normal proves this last one).
Credits to @NielsDiepeveen for noting this argument, in its cardinal function guise: $l(X) \ge |X|$ for locally countable spaces, and $l(X) \le d(X)$ for metacompact spaces. We don't have $l(X) = \mathfrak{c} \le d(X) = \aleph_0$.
$l(X) = \mathfrak{c}$ is witnessed by the closed and discrete irrationals.
$X$ is countably metacompact
$X$ is countably metacompact by a simple argument due to @NielsDiepeveen again: let $\mathcal{U} = \{U_n: n \in \mathbb{N}\}$ be a countable open cover of $X$. 
Then first shrink it to a countable cover $\{V_n: n \in \mathbb{N}\}$ with all $V_n$ open and $V_n \subseteq U_n$ such that $V_n \cap V_m \cap \mathbb{R\setminus Q} = \emptyset$ for $n \neq m$. (so that the open sets are disjoint on the irrationals).
This can be done as all subsets of $\mathbb{R\setminus Q}$ are closed in $X$, so 
we just define $$V_n = U_n \setminus \bigcup_{i < n} \left((\mathbb{R \setminus Q}) \cap U_i\right)$$ 
to get what we want. Every irrational is still in some $V_n$, because it won't be removed from the $U_n$ with the lowest index it appears in, and the rationals are unaffected.
Let $\{r_n: n \in \mathbb{N}\}$ be an enumeration of $\mathbb{Q}$, and define $W_n = X\setminus \{r_1, \ldots, r_n\}$ which are all open, as complements of finite sets.
Then define 
$$\mathcal{W} = \{ W_n \cap V_n: n \in \mathbb{Q}\} \cup \{\{q\}: q \in \mathbb{Q}\}\text{.}$$ 


*

*$\mathcal{W}$ is an open cover (any $x \in \mathbb{Q}$ is by definition covered, and if $x \notin \mathbb{Q}$ it's in some $V_n$ and so also in $W_n \cap V_n$ as this intersection only removes rationals).

*It's clear that $\mathcal{W} \prec \mathcal{U}$: any $\{x\}$ with $x$ rational is refined by any member of $\mathcal{U}$ that contains $x$. Any $W_n \cap V_n$ is refined by $V_n$ and so $U_n$.

*$\mathcal{W}$ is point finite: if $x \in \mathbb{R\setminus Q}$, it is in a unique $V_n$ and thus in a unique $V_n \cap W_n \in \mathcal{W}$, and if $x \in \mathbb{Q}$, say $x=r_k$ for some $k$, then the only members of $\mathcal{W}$ that $x$ could be in, are $\{x\}$ and $V_i \cap W_i$ for $i \le k$, so finitely many.
So $X$ is countably metacompact.
$X$ is not countably paracompact
As to countable paracompactness; a result (lemma 2.4) in

Fleissner, William. "Separation properties in Moore spaces". Fundamenta Mathematicae 98.3 (1978): 279-286. 

implies that $X$ is also never countably paracompact, I'll give the argument (written out with full details) for completeness (and because it's fun and educational, with thanks to professor Fleissner to clear up a question about the proof that he could elucidate): 
We'll have to use some set theory; the proof will be somewhat indirect, by diagonalisation. First note that we can use the regular open sets $O$ of $X$ (sets that obey $O = \operatorname{int}(\overline{O})$ as a base for the topology; this follows from regularity of $X$ and as $O \to O \cap \mathbb{Q}$ is a 1-1 map from all regular open sets to subsets of $\mathbb{Q}$, this base has size at most $\mathfrak{c}$)
In particular there are only (at most) $\mathfrak{c}^{\aleph_0} = \mathfrak{c}$ many sequences of regular open sets that form a locally finite cover of $X$, so enumerate them in type $\mathfrak{c}$ (long live AC, as usual in general topology):
Denote by $\{\mathcal{U}_\alpha: \alpha < \mathfrak{c}\}$ be an enumeration of all countable sequences of regular open sets that are locally finite covers of $X$. 
So write, for all $\alpha < \mathfrak{c}$: $$\mathcal{U}_{\alpha} = (U_{\alpha, i})_{i \in \omega}$$ These will be the candidate refinements for any open cover, and we enumerate the irrationals $\mathbb{P}$ (which are a closed and discrete subset of $X$) as $\mathbb{P} = \{y_\alpha: \alpha < \mathfrak{c}\}$ ($|\mathbb{P}| = |\mathbb{R}| = \mathfrak{c}$), and we are going to use $y_\alpha$ to "kill" $\mathcal{U}_\alpha$ for all $\alpha$.
To do this, for all $\alpha < \mathfrak{c}$, consider $y_\alpha$ and $\mathcal{U}_\alpha$; as the latter cover-sequence is locally finite (so point-finite) there are $i$ such that $y_\alpha \notin U_{\alpha,i}$, and define $n_\alpha$ to be the minimum of those $i$ (for definiteness).
Then define for $i \in \omega$: $$\mathbb{P}_i = \{y_\alpha \in \mathbb{P}: n_\alpha = i\}$$
and note that these form a partition of $\mathbb{P}$.
Then define $$V_n = X\setminus (\mathbb{P}\setminus\mathbb{P}_n), n \in \omega $$
And note these sets are open (as all subsets of $\mathbb{P}$ are closed in $X$)
The $V_n, n \in \omega$ also cover $X$: any $x \in \mathbb{Q}$ is in all of the $V_n$ and if we have $x \in \mathbb{P}$ so $x = y_\alpha$ for some $\alpha$, then $y_\alpha \in \mathbb{P}_{n_\alpha}$, so that $y_\alpha \notin \mathbb{P}\setminus \mathbb{P}_{n(\alpha)}$ and hence $x=y_\alpha \in V_{n_\alpha}$. In fact: 

$y_\alpha \in V_n$ iff $n_\alpha = n$, so that every point of $\mathbb{P}$ is in a unique $V_n$.

Now suppose, for a contradiction, that the countable open cover $\{V_n: n \in \omega\}$ has a locally finite refinement $\mathcal{W}$.
By a standard fact about locally finite refinements (see lemma 3.1 here) we can even say that there is a locally finite cover $\{W_n: n \in \omega\}$ such that for all $n$: $W_n \subseteq V_n$, and note that the collection $\{\operatorname{int} (\overline{W_n}) : n \in \omega\}$ is a locally finite regular open cover (as can easily be checked) and so it must have been one of our originally enumerated sequences , hence there is some $\beta < \mathfrak{c}$ such that
$$\forall n \in \omega: \operatorname{int}(\overline{W_n}) = U_{\beta, n}$$
But now where can $y_\beta$ be? It is in some $W_n$ (as these form a cover), and in fact by the property of the $V_n$ noted above,  it can only be in $W_{n_\beta}$, from which it follows that $y_\beta \in \operatorname{int}(\overline{W_n}) = U_{\beta, n_\beta}$, while $n_\beta$ was chosen so that $y_\beta \notin U_{\beta, n_\beta}$.
This is the required contradiction. 
So the cover has indeed been "killed", and so $X$ is not countably paracompact.
This is regardless of the choice of sequences in the definition of $X$, we just used that $\mathbb{P}$ is closed and discrete, and that there weren't too many open sequences to consider, so we could both enumerate them in the same type.
similar spaces
I think we can also simularly prove that spaces like Mrówka's $\Psi$-space (see here or here) will be countably metacompact and not countably paracompact, besides being a nice example of a non-normal pseudocompact but not countably compact space.
It has the same global structure of a large closed discrete set glued together with countably many dense isolated points. Even the Niemytzki-plane (Moore-plane) might be made to work, as we have a second countable space outside of the large discrete subspace, which could be sufficient to adapt the countably metacompactness proof.
At least the last proof goes through essentially unchanged for both of them, and all these spaces including our $X$, are non-normal by Jones' lemma.
