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
The proof you have given is correct, but the identical proof has been given on this site too. E.g. this one is very close. Or this one. Or this. It's the most "obvious" proof.