Here’s a very rough sketch of $X$:
V W --------(===)o(===)------------ --------(====x====)------------ a U b
The top line represents $C_2$ and the bottom line $C_1$. The diagram shows a basic open nbhd of $\langle x,1\rangle\in C_1$: the nbhd consists of the open interval $U=\big(\langle a,1\rangle,\langle b,1\rangle\big)$ about $\langle x,1\rangle$ in $C_1$ together with the two open intervals $V=\big(\langle a,2\rangle,\langle x,2\rangle\big)$ and $W=\big(\langle x,2\rangle,\langle b,2\rangle\big)$ in $C_2$. Points of $C_2$, on the other hand, are isolated.
Now let $Y$ be the result of identifying $C_1$ to a point $c^*$, and let $q:X\to Y$ be the quotient map, so that $q(x)=c^*$ for $x\in C_1$ and $q(x)=x$ for $x\in C_2$. Points of $C_2$ are still isolated. Suppose that $c^*\in H\subseteq Y$; then $H$ is open iff $q^{-1}[H]=C_1\cup\big(H\setminus\{c^*\}\big)$ is open in $X$. Thus, we want to know the answer to the following question:
For what sets $A\subseteq[0,1]$ is $C_1\cup\big( A\times\{2\}\big)$ open in $X$?
Suppose that $C_1\cup\big( A\times\{2\}\big)$ is open. Then for each $x\in[0,1]$ there must be an open interval $(a,b)$ such that $x\in(a,b)$ and $\Big((a,x)\cup(x,b)\Big)\cap[0,1]\subseteq A\;,$
to ensure that $C_1\cup\big( A\times\{2\}\big)$ contains the basic open nbhd $\left(\Big((a,b)\times\{1\}\Big)\cup\Big(\big((a,x)\cup(x,b)\times\{2\}\Big)\right)\cap X$ of $\langle x,1\rangle$.
This can be restated as follows: for each $x\in(0,1)$ there are $a,b\in(0,1)$ such that $x\in(a,b)$ and $A\supseteq(a,b)\setminus\{x\}$, and there are also $a,b\in(0,1)$ such that $A\supseteq(0,a)$ and $A\supseteq(b,1)$. In even simpler terms, each point of $(0,1)$ has an open nbhd in $[0,1]$ that contains at most one point of $[0,1]\setminus A$.
Added: To see this, suppose that $x\in(0,1)$. (The argument for $0$ and $1$ is very similar.) By the previous sentence there are $a,b\in(0,1)$ such that $x\in(a,b)$ and $A\supseteq(a,b)\setminus\{x\}$. If $x\in A$, then $A\supseteq(a,b)$, and $(a,b)$ is an open nbhd of $x$ in $[0,1]$ that is disjoint from $[0,1]\setminus A$. If $x\notin A$, then $(a,b)$ is still an open nbhd of $x$ in $[0,1]$, and the only point of $[0,1]\setminus A$ in $(a,b)$ is $x$ itself. Thus, in either case $(a,b)$ contains at most one point of $[0,1]\setminus A$.
But that just says that $[0,1]\setminus A$ is a closed, discrete set in $[0,1]$, which means that $[0,1]\setminus A$ must be finite.
In other words, if $c^*\in H\subseteq Y$, then $H$ is open in $Y$ iff $H$ contains all but finitely many points of $C_2$. Let $\mathscr{B}$ be the collection of all such subsets of $Y$; it’s easy to show that if $\mathscr{C}$ is a countable subset of $\mathscr{B}$, there is a $B\in\mathscr{B}$ that does not contain any $C\in\mathscr{C}$, so $\mathscr{C}$ cannot be a local base at $c^*$. Thus, $Y$ is not first countable at $c^*$.