If a topological space can be partitioned into finitely many second countable subspaces, is it second countable? [Collecting examples] A topological space is second countable if its topology has a countable basis. 
Let $(X,\tau)$ be a topological space. 

Suppose there exists $S_1,\ldots, S_n$ is a finite partition of $X$ such that, for each $1\leq i \leq n$, the subspace $(S_i,\tau|_{S_i})$ is second countable. Does it follow that $(X,\tau)$ is second countable? 

A negative answer to this question has already been given, but I'd be interested knowing more counterexamples. 
David Hartley has suggested this follow-up question:

If $(X,\tau)$ is first countable and it can be partitioned into finitely many second countable subspaces, is $(X,\tau)$ second countable?

 A: Let X be the result of glueing countably infinitely many copies of [0,1] together at 0. It is not second countable, it even fails to be first countable at 0. But {0} and X - {0} are each second countable.
(ETA) Suppose Y is an uncountable, second countable, T1 space and Z a countable, discrete space, with Y and Z disjoint. Let X be their union with the topology with a basis comprising the open sets of Y and sets of the form C∪{z} with z∈Z and C a cofinite subset of Y. X is not first countable as no point of Z has a countable neighbourhood basis in X, but it splits into Y and Z which are both second countable.
Both examples depend on X being not first countable, which suggests the follow-up question: If a first countable space can be partitioned into two second countable subspaces, is it second countable?
A: In General Topology by R.Engelking this example is given to show that the image $X_{/E}$ of a quotient mapping $f:X\to X_{/E}$ with closed equivalence classes  can fail to be $1$st-countable even when $X$ is $2$nd-countable:
Let $X=\mathbb R$ have the usual topology. Take  $p\not \in X.$ For $x,y \in \mathbb R$ let $xEy$ iff $(x=y\lor \{x,y\}\subset Z).$  Let $Y=(X$ \ $\mathbb Z) \cup \{p\}$. The quotient map $f:X\to Y$,  where $f(x)=x$ if $x\not \in Z$ and $f(x)=p$ if $x\in \mathbb Z,$ is called the identification of $\mathbb Z$ to a point.
The sub-space topology on $S_1= Y$  \ $\{p\} =\mathbb R$ \ $\mathbb Z$  is just the usual topology on $\mathbb R$ \ $\mathbb Z,$ which is $2$nd-countable. And the sub-space $S_2=\{p\}$ is (trivially) $2$nd-countable .
To show that the character of $p$ in $Y$ is uncountable, let $\{U_m:m\in \mathbb Z\}$ be a family of nbhds of $p.$  For each $ m$ take  $f_m:\mathbb Z\to (0,1/2]$ such that $$U_m\supset \{p\}\cup   \{(-f_m(n)+n,f_m(n)+n)\;):n\in \mathbb Z\} \;\backslash \;\mathbb Z.$$  Let $g(n)= f_n(n)/3$ for each $n\in \mathbb Z.$  Then $$V=\{p\}\cup \{(-g(n)+n,g(n)+n)\;):n\in \mathbb Z\}\; \backslash \; \mathbb Z$$ is a nbhd of $p.$ And $U_m\not \subset V$ for any $m\in \mathbb Z$ because $m+ 2f_m(m)/3 \in U_m$ \ $V$. 
