Let $X$ be a topological space and $\mathbb{U}$ an open cover of $X$.
(1) Suppose each $U$ $\in$ $\mathbb{U}$ contains a basis. Show that the union of all those bases is a basis for $X$.
(2) Show that if $\mathbb{U}$ is countable and each $U\in \mathbb{U}$ is countable then $X$ is second countable.
My attempt:
1) Let $U\in \mathbb{U}$. Define $\mathbb{B}$ $=$ $\{$ $\bigcup_{B \in \mathbb{B}_u}$ $B$ $:$ $\mathbb{B}_u$ is a basis for $\tau_{U}$ , where $U$ $\in \mathbb{U}$ $\}$. Note that if $A\in \mathbb{B}$ then it is open in $U$. Since $U$ is open in $X$, it follows that $A$ is open in $X$. Hence $\mathbb{B}$ is a collection of open subsets of $X$. Let $A$ be an open subset of $X$. Since $A\subseteq X$ and $\mathbb{U}$ covers $X$, it follows that $A\subseteq U_0$ for some $U_0 \in \mathbb{U}$. Let $\mathbb{B}_{U_0}$ be a basis for $U_0$, so $A\subseteq$ $\bigcup_{B\in \mathbb{B}_{U_0}}B$. This means that $A$ is contained within a basis element of $U_0$. Hence $A$ is the union of all possible basis elements of $U_0$. So $\mathbb{B}$ is a basis for $X$.
(2) If $\mathbb{U}$ is countable and each $U \in \mathbb{U}$ is countable then so are the basis elements of $U$, as they are subsets of $U$. So their unions is countable by cantor's theorem hence each $\mathbb{B_u}$ is countable for each $U$, since there are countably many such $U$, $\mathbb{B}$ is countable.
Is it correct? What would be an alternative proof?
The first part starts out OK: Suppose $\Bbb B_u$ is a base for $u \in \Bbb U$, then $$\Bbb B:= \bigcup \{\Bbb B_u: u \in \Bbb U\}$$
is indeed a collection of open sets of $X$, as a relatively open subset of an open subspace is open in the whole space.
But if $U \subseteq X$ is open, we cannot say that $U \subseteq U_0$ for some $U_0 \in \Bbb U$.
Better idea: reason by point:
Let $x \in U$, then $x \in u$ for some $u \in \Bbb U$ (each point is covered) As $\Bbb B_u$ is a base for $u$ and $x \in U \cap u$ (which is open in $u$) there is some $b \in \Bbb B_u (\subseteq \Bbb B)$ such that $x \in b \subseteq U \cap u$. It follows we have $b \in \Bbb B$ such that $x \in b \subseteq U$ and as $x \in U$ was arbitrary, $\Bbb B$ is a base for $X$.
Alternatively:
$U$ open in $X$ then for each $u \in \Bbb B$ the set $u \cap U$ is open in $u$ (possibly empty, but who cares) and so is a union of some subcollection of the base $\Bbb B_u$:
$$u \cap U = \bigcup U'_u \text{ for some } U'_u \subseteq \Bbb B_u$$
and then
$$U = U \cap X = U \cap \bigcup \Bbb U = \bigcup \{U \cap u: u \in \Bbb U\}= \bigcup_{u \in \Bbb U} \bigcup U'_u = \bigcup (\bigcup_u U'_u)$$
so $U$ is a union of elements of $\Bbb B$.
As to 2: this indeed follows from 1 and the fact that a countable union of countable sets is countable.