Basis for a topology with a countable number of sets I am working on the problems from the textbook "Topology without tears". I am stuck with problem number $4$ in Exercise $2.2$. Could anyone suggest some hints on how to proceed? The question goes as follows.
A topological space $(X,\tau)$ is said to satisfy the second axiom of countability or to be second countable if there exists a basis $\mathcal{B}$ for $\tau$, where $\mathcal{B}$ consists of only a countable number of sets.


*

*Prove that the discrete topology on an uncountable set does not satisfy the second axiom of countability.

*Let $(X,\tau)$ be the set of all integers with the finite-closed topology. Does the space $(X,\tau)$ satisfy the second axiom of countability?
For the first problem, I tried to argue by contradiction assuming that the basis for the topology is countable. But I did-not know what I need to look at to prove the contradiction. For the second one, I do not know where to start. I have not thought deeply on the second one though.
Thanks, Adhvaitha
 A: The first question can be proved directly: 
Since for every $x$ we have that $\{x\}$ is open, and in every basis for the topology every open set should be a union of basic open sets, the singletons cannot be anything but the union of themselves. Therefore every basis for the topology must contain $\Big\{\{x\}\mid x\in X\Big\}$ which is uncountable since it has the same cardinality of $X$ itself.
The second question can be easily solved with a theorem from elementary set theory:
Theorem: Suppose $X$ is a countable set, then $\{B\subseteq X\mid B\text{ finite}\}$ is also countable.
Proof:
Fix some enumeration of $X=\{x_n\mid n\in\mathbb N\}$. Suppose $A\subseteq X$ is a finite set, define: $$f(A) = \sum_{x_i\in A} 2^i$$
(In the $\mathbb N$ case, encode every finite set as a binary number)
The proof that this function is a bijection is done by strong (complete) induction, and is not difficult.

Now we can proceed with the second question. From the theorem, since the integers are countable, there are only countably many finite subsets of integers.
Since every set is closed if and only if it is finite, there are countably many finite closed sets, and therefore finitely many open sets (the complements of closed sets).
So the entire topology is a countable basis for itself, as needed.
A: For the point one, my hint is to think about the base
$$\mathcal{B}_0:=\{\{x\}|x\in X\}$$
and thinking in its relation to all the other posible basis of the discrete topology on a set $X$. If $\mathcal{B}$ is a basis for the discrete topology, what is its relation with $\mathcal{B}_0$? (See that for each $x\in X$, $\{x\}$ is open, and there is a nonempty open set $U\in\mathcal{B}$ such that $x\in U\subseteq \{x\}$.) And what is the cardinality of $\mathcal{B}_0$? (Clearly, $\mathcal{B}_0$ is equipotent to $X$.)
For the point two, I recommend you to think wether the family of finite subsets of a denumerable set is denumerable or not. Once you know that answer, the answer to point two would be inmmediate.
For proving that, my clue is to center on finite sets of order $n$. And how there is an injective map from the set of finite sets of order $n$ to the cartesian product of $X$ with itself $n$ times.
For more details or a more concrete form of the answer, say and I will give to you.
