Given $\mathbb{R}$ with the standard topology and $\{1,2\}$ with the discrete topology, consider the quotient $P:=\left(\mathbb{R}\times\{1,2\}\right)/\sim$ endowed with the quotient topology, where $(x,u)\sim(y,v)\Leftrightarrow x=y$.
Proposition: $P$ is second-countable.
Proof: First we realise that the product topology is $\left\{(a\dots b)\times\{1,2\}|a,b\in\mathbb{R},a<b\right\}$, where $(a\dots b)\subset\mathbb{R}$ s an open interval. For each $x\in\mathbb{R}$ the equivalence classes each have two elements because $\left[(x,1)\right]=\left[(x,2)\right]$. The quotient map is $\pi:\mathbb{R}\times\{1,2\}\rightarrow P,(x,u)\mapsto[(x,u)]$. We can see right ahead, that obviously $P\cong\mathbb{R}$, thus we write $[(x,u)]=x$.
The rational intervals $\{(q\dots r)|q,r\in\mathbb{Q}\}$ form a countable base of $\mathbb{R}$ and the point sets $\{1\},\{2\}$ form a countable base of the discrete space $\{1,2\}$. We can map them to $P$ under $\pi$, where we obtain $\pi\left((q\dots r)\times\{1\}\right)=\pi\left((q\dots r)\times\{2\}\right)=\{[x]|x\in\mathbb{Q},q<x<r\}$ as the open sets of $P$ endowed with the quotient topology. Because $P\cong\mathbb{R}$, this set is isomorphic to the set $\{(q\dots r)|q,r\in\mathbb{Q}\}$ of rational intervals. Thus, $P$ is second-countable, q.e.d..
Is this proof accurate?
Proof: First realise that the equivalence classes each have two elements: $(x,1);(x,2)$. We therefore use the notation $[x]:=[(x,1)]=[(x,2)]$. We then realise that the open sets of the quotient topology on $P$ are of the form $[B_\epsilon(x)]$. Now we can show that $P$ endowed with that topology is homeomorphic to $\mathbb{R}$. We map the open sets in $P$ onto the open sets in $\mathbb{R}$: $[B_\epsilon(x)]\mapsto B_\epsilon(x)$. It is obvious that this is a bijection and continuous. Thus, we have $\mathbb{R}\cong P$ and since second countability is a topological invariant, $P$ is second countable because $\mathbb{R}$ is, q.e.d..