I realize this question has been asked, But i just wrote up this proof and it's a little different so I was hoping somebody would check it.
Let $f,g: X \rightarrow Y$ be continuous and $Y$ be Hausdorff. Prove that $Z=\{x | f(x) = g(x)\}$ is a closed subset of $X$.
Proof: Let $x \in X-Z$
Then $f(x) \neq g(x)$, and since $Y$ is hausdorff this implies that $\exists U_1, U_2$ open in $Y$ s.t. $f(x) \in U_1$, $g(x) \in U_2$ and $U_1 \cap U_2 = \emptyset$.
Since $f$ is continuous and $U_1$ is open in $Y$, we have $f^{-1}(U_1)$ is open in $X$ and $x \in f^{-1}(U_1)$.
Since $g$ is continuous and $U_2$ is open in $Y$, we have $g^{-1}(U_2)$ is open in $X$ and $x \in g^{-1}(U_2)$.
Claim: $g^{-1}(U_2) \cap f^{-1}(U_1)$ is a neighborhood of $x$ contained in $X-Z$.
Suppose $z \in g^{-1}(U_2) \cap f^{-1}(U_1)$, then $f(z) \in U_1$, $g(z) \in U_2$. But $U_1 \cap U_2 = \emptyset$ and so $f(z) \neq g(z)$ and thus $z \in X-Z$.
Thus $X-Z$ is open and thus $Z$ is closed
