If $f,g:X \to Y$ are continuous and $Y$ is $T_2$, then $\{x \in X\,|\,f(x)=g(x)\}$ is closed I'd like to know if the following proof is valid.  The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.
"Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions.  Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."
Let $x \in X$ be such that $f(x) \ne g(x)$.  Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$.  Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$.  Then $U \cap f(C)=V \cap f(C)=\varnothing$.  Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$.  Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.
Thanks.
 A: Different argument (assuming you did the exercise in the link before):  
If $f,g:X\to Y$ are continuous and $Y$ is Hausdorff then by a closed graph theorem we have that $G_f=\{(x,f(x)):x\in X\}$ and $G_g=\{(x,g(x)) : x\in X\}$ are closed in $X\times Y$; therefore, $G_g\cap G_f = \{(x,f(x) ):x\in X , f(x)=g(x)\}$ is closed as well.  But $F:X\to X\times Y$ given by $F(x)=(x,f(x))$ is continuous , whence $F^{-1}(G_f\cap G_g)=\{x : f(x)=g(x)\}$ is closed in $X$.
A: If $f,g \colon X \to Y$ are continuous, then $(f,g)\colon X\to Y\times Y$ is continuous. The set $\{x\colon f(x) = g(x) \}$ is the counter-image by means of $(f,g)$ of the diagonal $\Delta = \{ (y,y) \colon y\in Y\}$ of $Y\times Y$. So it is enough to check that the diagonal is a closed set. 
Let's prove that any point $(x,y)\not \in \Delta$ has a neighborhood which does not intersect $\Delta$. In fact Hausdorff property of $Y$ states that $x$ and $y$ (being different points) have two non overlapping neighbourhoods $U$, $V$. Hence $U\times V$ is a neighbourhood of $(x,y)$ not touching $\Delta$.
