Is every regular, separable, first countable, pseudocompact, linearly Lindelöf Moore space Lindelöf? Suppose that $X$ is regular, separable, first countable, pseudocompact, linearly Lindelöf and it is a Moore space. Is it Lindelöf?

$X$ is linearly Lindelöf if every open cover of $X$ that is a chain with respect to $\subseteq$ has a countable subcover.

 A: It is easy to check that each linearly Lindelöf space has countable extent. We recall that a space $X$ is a $\sigma$-space if $X$ has a $\sigma$-discrete network. [Gru, 4.3] Every Moore space $X$ is a $\sigma$-space. [Gru, 4.5] Moreover, a space $X$ is a Moore space iff $X$ is a $\sigma$-space and a $p$-space (or a $w\Delta$-space). [Gru, 4.7.(i)]. If $X$ is a $\sigma$-space and $e(X)=\omega$ then $nw(X)=\omega$, see the proof. So each linearly Lindelöf $\sigma$-space has countable network, and, therefore, it is Lindelöf, which already answers positively your question. But we shall continue. If $X$ is a $p$-space, then $nw(X)=w(X)$. [Gru, 4.2] In particular, each linearly Lindelöf Moore space is a second countable regular space. By Nagata-Smirnov Theorem [Eng, 4.4.7], a topological space $X$ is metrizable iff $X$ is regular and has a $\sigma$-locally finite base. In particular, each linearly Lindelöf Moore space is a second countable regular space and is metrizable. (Another proof of this fact is the following: Let $X$ be a linearly Lindelöf Moore space. Then, by the above, the space $X$ is Lindelöf. By [Eng, Theorem 5.1.2], each regular Lindelöf space is paracompact. By [Gru, Cor. 3.4] a paracompact Moore space is metrizable. Clearly, each metrizable Lindelöf space is second countable). Next, if $X$ is a linearly Lindelöf Moore pseudocompact space then $X$ is a metrizable compact. Indeed, by the above, the space $X$ metrizable, and, therefore, normal. By [Eng, Th. 3.10.21] every pseudocompact normal space is countably compact. And a countably compact Lindelöf space is compact. So the space from your question is a metrizable compact. 
Remark 1. See Propositions 1.2.(7) and Remark 1.3 from [BR] about similar relations between cardinal invariants of Moore spaces. Which I, after writing this answer, think to rewrite, because it seems that $e(X)=w(X)$ for each Moore space $X$ and, moreover, if $e(X)=\omega$ then the space $X$ is metrizable, so $dc(X)=w(X)=\omega$. 
Remark 2. Under CH there is an other approach for you question, because this assumption implies that each separable linearly Lindelöf regular space is second countable. Indeed, it is easy to prove that any open cover of cardinality $\omega_1$ of a linearly Lindelöf space has a countable subcover. By easily proved Theorem 1.5.7 from [Eng], $w(X)\le 2^{d(X)}$ for each regular space $X$. In particular, each separable linearly Lindelöf regular space is Lindelöf.
Remark 3. A space $X$ is subparacompact if every open cover of $X$ has a $\sigma$-discrete closed refinement. 
See Sections 3 and 4 of [Bur] about definitions of and relationships among covering properties. 
Proposition.  If $X$ is a subparacompact Hausdorff space then $e(X)=l(X)$.
Proof. Since $e(X)\le l(X)$, it suffices to prove that $l(X)\le e(X)$. Fix an open cover $\mathcal U$ of $X$. By Theorem 3.1 of [Bur], there is a sequence $\{\mathcal G_n\}$ of open refinements such that for any point $x\in X$ there exists $n$ such that $\operatorname{ord} (x,\mathcal G_n)=1$. For each $n$ put $X_n=\{x\in X: \operatorname{ord} (x,\mathcal G_n)=1\}$ and $\mathcal H_n=\{U\in\mathcal G_n: X_n\cap U\ne\varnothing\}$. For each element $U\in\mathcal H_n$ pick an arbitrary point $x(U)\in U\cap X_n$ and put $Y_n=\{x(U):U\in\mathcal H_n\}$. Since $Y_n\subset X_n$, the map $x:\mathcal H_n\to Y_n$ is injective, so $|\mathcal H_n|\le |Y_n|$. We claim that $Y_n$ is a closed and discrete subset of the space $X$, so $|Y_n|\le e(X)$. Indeed, let $x\in X$ be an arbitrary point. Pick an arbitrary member $U$ of the family $\mathcal G_n$ which covers the point $x$. Let $y\in U\cap Y_n$ be an arbitrary point. Since $y\in X_n$,  $\operatorname{ord} (y,\mathcal G_n)=1$, hence $y=x(U)$, that is there is at most one such a point $y$. Next, $X_n\subset\bigcup\mathcal H_n$. Since $\mathcal H_n\subset \mathcal G_n$ and $\mathcal G_n$ is a refinement of the cover $\mathcal U$, there exists a subcover $\mathcal V_n$ of $\mathcal U$ of cardinality at most $|\mathcal H_n|\le |Y_n|$, which covers the set $X_n$. Put $\mathcal V=\bigcup\mathcal V_n$. Then  $\mathcal V\subset\mathcal U$ is a cover of the set $\bigcup X_n=X$ and $|\mathcal V|\le \sum|\mathcal V_n|\le\omega\cdot e(X)=e(X)$.
Corollary 1. Let $X$ be a regular space. If $X$ is Moore, or paracompact, or  a $\sigma$-space, or a strong $\Sigma$-space then $e(X)=l(X)$. 
Proof. All these spaces are subparacompact, see [Gru].
Corollary 2. Each linearly Lindelöf subparacompact Hausdorff space is Lindelöf. 
References
[BR] Taras Banakh, Alex Ravsky. Verbal covering properties of topological spaces // arXiv: 1503.04480 [math.GN], Topology Appl. (Proceedings of Lepanto Conference) (to be published).
[Bur] Dennis K. Burke, Covering properties, p. 347-422 in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.
[Eng]  Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.
[Gru] Gary Gruenhage Generalized Metric Spaces, p. 423-501 in: K.Kunen, J.E.Vaughan (eds.) Handbook of Set-theoretic Topology, Elsevier Science Publishers B.V., 1984.
