Showing that a certain subset is Hausdorff I have a space $X$ with a subset, $A$, and then I let $Y = X \times \{0,1\} /  R$ where $R$ is the equivalence relation defined by the partition $\{(a,0), (a,1)\}$ if $a \in A$ and $\{(x,i)\}$ for $i$ equal to $0$ or $1$. ($\{0,1\}$ has the discrete topology).
Now I am trying to show that if $Y$ is Hausdorff then $X$ is Hausdorff and $A$ is closed. I am using the fact that the graph of $R$, say $G \subset Y^2$, is closed. Now points $((x,i),(y,j)) \in G$ iff $x=y$ and ($x \in A$ or $i=j$). I also know that if we give $Y^2$ the product topology then $Y^2 -G $ is a union of sets $U\times V$ where $U$ and $V$ are open in $X \times \{0,1\}$. 
But I am now really struggling to show that this gives me the required conclusion, I'm finding it hard to see what is actually going on with the graph of $R$. Thanks
 A: I would take a more pedestrian approach. Pick $x \in \overline{A}$. Let $V_i$ be a neighbourhood of $[(x,i)]$ in $Y$ (where $[(x,i)]$ denotes the equivalence class of $(x,i)$ modulo $R$). By the continuity of the projection, there are open neighbourhoods $U_i$ of $(x,i)$ in $X\times \{0,1\}$ that are mapped into $V_i$. Since $X\times\{i\}$ is open in $X\times\{0,1\}$, we may assume that $U_i$ is of the form $W_i \times\{i\}$ for an open neighbourhood $W_i$ of $x$. Let $W = W_0 \cap W_1$. Since $x \in \overline{A}$, there is an $a_W \in A\cap W$. But then $[(a,0)] = [(a,1)] \in \pi(W\times\{0\}) \cap \pi(W\times\{1\}) \subset V_0\cap V_1$, so $[(x,0)]$ and $[(x,1)]$ have no disjoint neighbourhoods, and since $Y$ is Hausdorff, that means $[(x,0)] = [(x,1)]$ or $x\in A$, hence $A$ is closed.
Now we show that the projection embeds $X\times \{0\}$ into $Y$, which then, since that space is homeomorphic to $X$, means $X$ is Hausdorff. Since the projection is continuous by definition of the quotient topology, and evidently injective on $X\times\{0\}$, it suffices to see that it is closed. Let $C \subset X$ be closed. Then $\pi^{-1}(\pi(C\times\{0\})) = C\times\{0\} \cup (C\cap A)\times\{1\}$. But $C\cap A$ is closed in $X$, so $\pi^{-1}(\pi(C\times\{0\}))$ is closed in $X\times\{0,1\}$, whence $\pi(C\times\{0\})$ is closed, and so $\pi\lvert_{X\times\{0\}}$ is closed, hence an embedding.
