$X \models S_f(\mathcal O,\mathcal O)$ implies $X \models S_1(\mathcal O,\mathcal O)$ I have a problem with a begining of a proof in a course that I am reading. I bring here the definitions and statement:
Definition (Rothberger): For $k \in \mathbb N$, a space $\langle X,O \rangle$ satisfies Rothberger ptoperty or $S_k(\mathcal O,\mathcal O)$, iff for any family of open covers of $X$, $\langle \mathcal U_n | n \in \mathbb N \rangle$ there exists some
$\langle \mathcal F_n \in [\mathcal U_n]^k | n \in \mathbb N \rangle$ such that $\bigcup _{n \in \mathbb N} \mathcal F_n$ covers $X$.
Observation: For a topological space $\langle X,O \rangle$, TFAE:
1.$X \models S_1(\mathcal O,\mathcal O)$
2.$X \models S_k(\mathcal O,\mathcal O)$ for some $k \in \mathbb N$
3.$X \models S_f(\mathcal O,\mathcal O)$ for some $f \in \mathbb N^{\mathbb N}$. i.e. for any family of open covers of $X$, $\langle \mathcal U_n | n \in \mathbb N \rangle$, 
There exists a family $\langle \mathcal F_n \in [\mathcal U_n]^{f(n)} | n \in \mathbb N \rangle$ such that $\bigcup_{n \in \mathbb N}  \mathcal F_n $ is an open cover of $X$.
Proof: To see $(c) \Rightarrow (a)$, fix $f \in \mathbb N^{\mathbb N}$, such that $X \models S_f(\mathcal O,\mathcal O)$. Pick an arbitrary partition $\langle A_n \in [\mathbb N]^{f(n)} \rangle$ with $\uplus_{n \in \mathbb N} A_n = \mathbb N$.
For all $n \in \mathbb N$, Let $\mathcal V_n := \{ \bigcap Im(g) | g \in \Pi_{m \in A_n} \mathcal U_m \}$. Evidently, each $\mathcal V_n$ covers $X$...
My question is: How can we conclude that each $\mathcal V_n$ covers $X$?
To me, it seems like, for each $n \in \mathbb N$, $\mathcal V_n$ is an open subset of $X$. What am I missing?
Thank you! 
 A: First, $\mathcal{V}_n$ consists of finite intersections of open subsets of $X$ (since $\mathrm{Im}(g)$ is a finite set of open subsets of $X$), so it is a collection of open sets (and not an open subset of $X$). 
Second, note that the particular $A_n$ and $\langle \mathcal{U}_m \mid m\in A_n \rangle$ are not important, so let's reduce the number of symbols we are dealing with. 
Fix a natural number $k$. For $1 \leq i \leq k$, let $\mathcal{U}_i$ be an open cover. We want to prove that $\{\bigcap \mathrm{Im}(g) \mid g \in \Pi_{1 \leq i \leq k} \mathcal{U}_i\}$ is an open cover of $X$. We do this by induction on $k$. The base case $k =1$ is clear.
For $1 \leq i < k$, fix $h \in \Pi_{1 \leq i < k} \mathcal{U}_i$ and let $V_h = \bigcap_{1 \leq i <k}h(i)$. Let $G_h = \{g \in \Pi_{1 \leq i \leq k}\mathcal{U}_i \mid 1 \leq i <k \implies g(i) = h(i)\}$.  
Then for $g \in G_h$, $\bigcap \mathrm{Im}(g) = V_h \cap g(k)$. So $\bigcup\{\bigcap \mathrm{Im}(g) \mid g \in G_h\} = \bigcup \{V \cap U\mid U \in \mathcal{U}_k\} = V_h $.
So, $\bigcup \{\bigcap \mathrm{Im}(g) \mid g \in \Pi_{1 \leq i \leq k} \mathcal{U}_i\} = \bigcup\{\bigcup\{\bigcap \mathrm{Im}(g) \mid g \in G_h\} \mid {h \in \Pi_{1 \leq i < k}}\mathcal{U}_i\} = \bigcup \{V_h \mid {h \in \Pi_{1 \leq i < k}} \mathcal{U}_i\}$. 
And this is equal to $X$ by the inductive hypothesis.
So each of the $\mathcal{V}_n$ in the question are open covers of $X$.
