Countably local finiteness and a related(?) property I'd like to know how the following properties are related:
1.) $\mathcal{O}$ is a cover of $X$ such that every point of $X$ has a neighborhood that intersects at most countably many members of $\mathcal{O}$
2.) $\mathcal{O}$ is countably locally finite.
Is there a simple relationship between the two, e.g. 1 implies 2 or 2 implies that some subcover satisfies 1?  Is their relationship strengthened in the presence of any mild assumptions, e.g. $X$ is Hausdorff, etc.?
Background/what I've tried: I thought about this while trying to get a better understanding of countably local finiteness.  It seemed to me that 1 was a simple generalization for local finiteness, and it could serve as a nice heuristic if the two were equivalent.  I started trying to show that they were related (or, better, equivalent), but I kept running into problems in proving it.  I tried taking a cover satisfying 1 and breaking it into a bunch of locally finite covers, but I couldn't find a systematic way of doing so without worrying about the affects that a choice of decomposition at a single point would have on neighboring points.  Maybe choice can patch that up?  And with the reverse implication, given a countably locally finite cover, my problem was that for a given point, there's a nice neighborhood that corresponds to each of my locally finite subcollections, but I couldn't find a guarantee that any of those nice neighborhoods would intersect at most countably many sets in the whole cover...
Any words of wisdom?
 A: A locally countable open cover need not be $\sigma$-locally finite.
For this consider the open ordinal space $\omega_1 = [ 0 , \omega_1 )$ with the usual order topology, and the open cover $\mathcal{U} = \{ [ 0 , \omega_1 ) \} \cup \{ ( \alpha , \omega_1 ) : \alpha < \omega_1 \}$. It is pretty easy to see that this is locally countable, as each point of $\omega_1$ has a countable neighbourhood, and each point is an element of countably many sets in $\mathcal{U}$.  But this cover is not $\sigma$-locally finite (even not $\sigma$-point finite). If it were partitioned as $\mathcal{U} = \bigcup_{i \in \mathbb{N}} \mathcal{U}_i$, then some $\mathcal{U}_i$ must be infinite. Taking any $\beta \in \omega_1$ such that $( \alpha , \omega_1 ) \in \mathcal{U}_i$ for infinitely many $\alpha < \beta$, it follows that $\beta$ is an element of infinitely many sets in $\mathcal{U}_i$.
(Of course, this open cover does have a $\sigma$-locally finite (indeed, simply finite) subcover.)
To go a bit further, recall that a (Hausdorff) space is paracompact if every open cover has a locally finite open refinement. Recall, too, the following theorem of Ernest Michael

Theorem. A regular space is paracompact iff every open cover has a $\sigma$-locally finite open refinement.

Now, a (Hausdorff) space is called para-Lindelöf if every open cover has a locally countable open refinement. One of the basic questions of this class of spaces was whether all regular para-Lindelöf spaces are paracompact.
In her PhD thesis, Caryn Navy showed that the answer to this question was negative: there are regular (even normal) para-Lindelöf spaces which are not paracompact. In terms of the properties given above, in such a space there is a locally countable open cover with no $\sigma$-locally finite open refinement.  This is a very strong rejection of any implication from (1) to (2).

A $\sigma$-locally finite open cover need not be locally countable.
Consider the space $X = \{ * \} \cup ( \omega \times \omega_1 )$ where the points of $\omega \times \omega_1$ are isolated, and the open neighbourhoods of $*$ are of the form $\{ * \} \cup A$ where $A \subseteq \omega \times \omega_1$ has the property that for co-finitely many $n \in \omega$ there are co-countably many $\alpha \in \omega_1$ such that $\langle n , \alpha \rangle \in A$. We consider the following cover of $X$: $$\mathcal{U} = \{ X \} \cup \{ \{ \langle n , \alpha \rangle \} : n \in \omega , \alpha \in \omega_1 \}.$$
It should be clear that every neighbourhood of $*$ meets uncountably many sets in $\mathcal{U}$.  However, $\mathcal{U}$ is $\sigma$-locally finite: setting $\mathcal{U}_n = \{ \{ \langle n , \alpha \rangle \} : \alpha \in \omega_1 \}$ and $\mathcal{U}_* = \{ X \}$, we have that $\mathcal{U} = \mathcal{U}_* \cup \bigcup_{n \in \omega} \mathcal{U}_n$ and that each $\mathcal{U}_n$ is locally finite. (Note that $V_n = \{ * \} \cup ( ( \omega \setminus \{ n \} ) \times \omega_1 )$ is an open neighbourhood of $*$ that is disjoint from every set in $\mathcal{U}_n$.) Note that this space is normal.
One should be able to come up with a $\sigma$-locally finite open cover with no locally countable subcover, but I have been as yet unable to think of an appropriate space/cover. As for refinements, I am quite uncertain.

The only "non-trivial" positive connection between the two that I can come up with is that if $X$ is a P-space (meaning that the family of open sets is closed under countable intersetions, or all Gδ subsets are open), then every $\sigma$-locally finite open cover of $X$ is locally countable.
