Solution: We will show that the complement $E^c$ is open in $X$. Let $x \in E^c$. Since $x \in E^c$, so $f(x) \neq g(x)$. As $Y$ is Hausdorff, we can find two disjoint open sets $V_{f(x)}$ and $V_{g_{x}}$ containing $f(x)$ and $g(x)$ respectively. Define the set \[ U = f^{-1} \left( V_{f(x)} \right) \cap g^{-1} \left( V_{g(x)} \right) . \] Since $f$ and $g$ are continuous, and intersection of two open sets is open, $U$ is open. We claim that $U \subseteq E^c$. Let \begin{align*} y \in U & \implies f^{-1} \left( V_{f(x)} \right) \cap g^{-1} \left( V_{g(x)} \right) \\ & \implies f(y) \in V_{f(x)} \text{ and } g(y) \in V_{g(x)}. \end{align*} Since $V_{f(x)} \cap V_{g(x)} = \emptyset$, so $f(y) \neq g(y)$ and hence $y \in E^c$. Thus, $U \subseteq E^c$ and hence, $E^c$ is open.