Are star countable spaces with $n$-in-countable bases Lindelöf? 
Claim: Let $X$ be a star countable space with an $n$-in-countable base for
  some $n \in N$. Then $X$ is Lindelöf.
Definition 1: A family $\mathcal U$ of subsets of a space $X$ is called $k$-in-countable if every set $A \subset X$ with $|A|=k$ is contained in at most countably many elements of $\mathcal U$.
Definition 2: A topological space $X$ is said to be star countable if whenever $\mathscr{U}$ is an open cover of $X$, there is a countable subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$. (Where $\operatorname{St}(K,\mathscr{U}) = \bigcup \{ U \in \mathscr{U} : K \cap U \neq \emptyset \}$.)

Thanks for your help.
 A: Proof: Let $\mathcal B$ be an $n$-in-countable base for some $n \in N$.
Suppose not. So there is an uncountable open cover $\mathcal U_0$ of
elements of $\mathcal B$ which has no countable subcover. Enumerate
it as $\{B_\alpha: \alpha < \kappa \}$. 
Since $X$ is star countable,
there is a countable subset $A_0$ such that $\operatorname{St}(A_0,
\mathcal U_0) = X$. Then there is some point $x_0 \in A_0$ such that
$\mathcal B_0 = \{B \in \mathcal U_0: x_0 \in B\}$ is uncountable
and $\mathcal B_0$ hasn't a countable subcover of $\bigcup \mathcal
B_0$, otherwise $\mathcal U_0$ has a countable subcover of $X$. Put
$\alpha_0= \min \{\alpha < \kappa: x_0 \in B_\alpha \}$. Let
$\mathcal V_0=\{B_\alpha \in \mathcal U_0: x_0 \in B_\alpha, \alpha
> \alpha_0\}$, $\mathcal U_1 = \{B_\alpha \setminus \{x_0\}:
B_\alpha \in \mathcal V_0\}$. Let $\mathcal W_1 = \mathcal U_1 \cup
(\mathcal U_0 \setminus \mathcal V_0) \cup \{B_{\alpha_0}\}$.
We can pick a countable subset $A_1$ such that $\operatorname{St}(A_1,
\mathcal W_1) = X$. Then there is some point $x_1 \in A_1$ such that
$\mathcal B_1 = \{B \setminus \{x_0\}\in \mathcal U_1: x_1 \in B
\setminus \{x_0\} \}$ is uncountable and $\mathcal B_1$ hasn't a
countable subcover of $\bigcup \mathcal B_1$, otherwise $\mathcal
B_0$ has a countable subcover of $\bigcup \mathcal B_0$.  Put
$\alpha_1= \min \{\alpha < \kappa: x_1 \in B_\alpha \setminus
\{x_0\} \in \mathcal U_1 \}$. Let $\mathcal V_1=\{B_\alpha \setminus
\{x_0\} \in \mathcal U_1: x_1 \in B_\alpha \setminus \{x_0\}, \alpha
> \alpha_1\}$, $\mathcal U_2 = \{B_\alpha \setminus \{x_0, x_1\}:
B_\alpha \in \mathcal V_1\}$. Let $\mathcal W_2 = \mathcal U_2 \cup
(\mathcal U_1 \setminus \mathcal V_1) \cup \{B_{\alpha_1}\} \cup
(\mathcal U_0 \setminus \mathcal V_0) \cup \{B_{\alpha_0}\}$.
Continuing in this way, for each $k \le n-1$, we can choose a
countable set $A_k$, a point $x_k$ of $A_k$, an uncountable family
$\mathcal U_k$ and an uncountable subset $\mathcal B_k = \{B
\setminus \{x_0,x_1,..., x_{k-1}\} \in \mathcal U_k : x_k \in B
\setminus \{x_0,x_1,..., x_{k-1}\} \}$. At the step $n-1$, we can
get a subset $\{x_0, x_1,..., x_{n-1}\}$ of $X$ such that  $\{x_0,
x_1,..., x_{n-1}\}$ is contained in uncountable many elements of
$\mathcal B$. 
This is a contradiction!
