Is there another method to characterize Lindelöf property? As the title, I want to know whether there is a method to characterize the Lindelöf property by the limit points, just as we characterize the countably compact.
 A: I think that if there were such a characterization, I’d probably have seen it at some point, but I don’t know of one. The obvious candidate,

every uncountable subset of $X$ has a cluster point (equivalently, $e(X)=\omega$, where $e(X)$ is the extent of $X$, the supremum of cardinalities of closed discrete subsets of $D$),

fails badly: $\omega_1$ is a very nice space (e.g., first countable, countably compact, and perfectly normal) with countable extent that is not Lindelöf.
A better candidate is the complete accumulation point condition:

(CAP): every uncountable subset of $X$ of regular cardinality has a complete accumulation point in $X$.

However, it turns out that this is equivalent not to $X$ being Lindelöf, but to $X$ having the weaker property of being linearly Lindelöf:

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

Proposition: A space $X$ satisfies the CAP iff $X$ is linearly Lindelöf.

Proof: Suppose first that $X$ is not linearly Lindelöf, and let $\mathscr{U}$ be an open cover of $X$ linearly ordered by $\subseteq$ that has no countable subcover. Without loss of generality we may assume that $\mathscr{U}=\{U_\xi:\xi<\kappa\}$ for some regular cardinal $\kappa$ with $\operatorname{cf}\kappa>\omega$ and that $U_\xi\subsetneqq U_{\xi+1}$ for each $\xi<\kappa$. For $\xi<\kappa$ choose $x_\xi\in U_{\xi+1}\setminus U_\xi$, and let $A=\{x_\xi:\xi<\kappa\}$. Clearly $A$ is uncountable, and it’s not hard to see that $A$ has no complete accumulation point: for any $x\in X$ there is a minimal $\xi(x)<\kappa$ such that $x\in U_{\xi(x)}$, and $|U_{\xi(x)}\cap A=|\xi(x)|<\kappa=|A|$.
Now suppose that there is an $A=\{x_\xi:\xi<\kappa\}\subseteq X$ of regular uncountable cardinality $\kappa$ that has no complete accumulation point. For each $x\in X$ let $U(x)$ be an open nbhd of $X$ such that $|U(x)\cap A|<\kappa$, and let $\alpha(x)=\min\big\{\eta<\kappa:U(x)\cap A\subseteq\{x_\xi:\xi<\eta\}\big\}$. Finally, for $\xi<\kappa$ let $V_\xi=\bigcup\{U(x):\alpha(x)=\xi\}$, and let $\mathscr{V}=\{V_\xi:\xi<\kappa\}$; $\mathscr{V}\,$ is clearly a linearly ordered open cover of $X$ with no countable subcover. $\dashv$

Obviously every Lindelöf space is linearly Lindelöf, but the converse is false. The simplest Tikhonov counterexample that I’ve seen is due independently to Gary Gruenhage and Raushan Buzyakova:

Example: Let $D=\{0,1\}$ have the discrete topology, and let $$X=\{x\in D^{\omega_\omega}:|\operatorname{supp}(x)|<\omega_\omega\}\;,$$ where $\operatorname{supp}(x)=\{\xi<\omega_\omega:x(\xi)=1\}$.
Proof: Let $A\subseteq X$ have regular uncountable cardinality $\kappa$. Clearly $D^{\omega_\omega}$ and hence $X$ has a base of cardinality $\omega_\omega$, so if $\kappa>\omega_\omega$, $A$ must have a complete accumulation point. If not, each $x\in X$ has a basic nbhd $U(x)$ such that $|U(x)\cap A|<\kappa$, and since $\kappa$ is regular this implies that $$|A|=\left|\bigcup_{x\in X}\big(U(x)\cap A\big)\right|<\kappa\;,$$ which is impossible.
Now suppose that $\kappa<\omega_\omega$. For $n\in\omega$ let $X_n=\{x\in X:|\operatorname{supp}(x)|\le\omega_n\}$, and let $A_n=A\cap X_n$. Since $\kappa>\omega$ is regular, there must be some $k\in\omega$ such that $|A_k|=\kappa$, and since $\kappa<\omega_\omega$, there is also an $n\in\omega$ such that $\kappa\le\omega_n$. Let $m=\max\{k,n\}$, let $$S=\bigcup_{x\in A_m}\operatorname{supp}(x)\;,$$ and let $H=\{x\in D^{\omega_\omega}:\operatorname{supp}(x)\subseteq S\}$; this is a closed subset of $D^{\omega_\omega}$, so it’s compact. On the other hand, $|S|\le|A_m|\cdot\omega_m=\kappa\cdot\omega_m=\omega_m<\omega_\omega$, so $H\subseteq X$, and since $A_m\subseteq H$, it follows immediately that $A_m$ has a complete accumulation point in $H$ and hence that $A$ has a complete accumulation point in $X$. This shows that $X$ is linearly Lindelöf.
To see that $X$ is not Lindelöf, note first that $Y=\{x\in D^{\omega_\omega}:|\operatorname{supp}(x)|\le\omega\}$ is a dense subset of $X$. $Y$ is the $\Sigma$-product of compact spaces and therefore is countably compact. It’s easy to show that a Tikhonov space with a dense countably compact subset is pseudocompact, and a pseudocompact Lindelöf space is compact, so if $X$ were Lindelöf, it would be compact. But $X$ is a dense proper subset of the compact space $D^{\omega_\omega}$, so $X$ is not compact and hence not Lindelöf. $\dashv$

Ken Kunen has even constructed a locally compact counterexample of weight $\omega_\omega$, which is the minimum possible weight for such a counterexample.
A: There is an answer from the theory of uniformities. My source here is the book Modern Analysis and Topology, chapter 3, by Norman R. Howes. (An open cover $\mathcal{U}$ is called normal whenever there is a sequence $\mathcal{V}_n$, $n \in \mathbb{N}$ of open covers of $X$ such that $\mathcal{V}_1$ refines $\mathcal{U}$ and for every $n$, $\mathcal{V}_{n+1}$ star-refines $\mathcal{V}_n$.) 
A theorem by Shirota says that for a completely regular space $X$ the set of countable normal covers is a base for a uniformity called $e$ (in the covering definition) that induces the original topology on $X$.
Then a theorem by Howes says that (a completely regular space) $X$ is Lindelöf iff every transfinite sequence that is cofinally Cauchy w.r.t the uniformity $e$ has a cluster point.
Writing out the definitions: every sequence $(x_\alpha)_{\alpha < \gamma}$ such that for every normal countable (open) cover $\mathcal{U}$ of $X$, there is a cofinal subset $A$ of $\gamma$ and a member $O$ of $\mathcal{U}$, such that $x_{\alpha} \in O$ for all $\alpha \in A$, there exists $p$ in $X$, such that every for every open neighborhood $U$ of $p$ we have a cofinal subset $B$ of $\gamma$ such that $x_{\alpha} \in U$ for all $\alpha \in B$.
This does give a characterization of Lindelöfness in terms of cluster points, but is not very nice to handle, and still involves covers of $X$, in the end.
I concur with Brian that I haven't seen such a characterization as well. The linearly Lindelöfness is as close as we get, and a countably metacompact linearly Lindelöf space is Lindelöf, so the notions are pretty close (as countable metacompactness is a relatively mild condition, as covering properties go).  
