Solution: Consider the map \[ \phi : X \to G, \quad x \mapsto (x, f(x)). \] Note that $\phi (x) = (Id \times f)(x,x)$, where $Id$ is the identity function on $X$. Since $Id$ and $f$ both are continuous functions, their product will be continuous and hence $\phi $ is continuous. Also, \begin{align*} h(x) = h(y) & \implies (x, f(x)) = (y, f(y)) \implies x = y. \end{align*} Thus, $h$ is injective. Now for the surjectivity, for any $(x,f(x)) \in G$, $x \in X$ and hence $\phi (x) = (x,f(x))$. Thus $\phi $ is a bijective continuous function from $X$ to $G$. Also, the function \[ \psi : G \to X, (x,f(x)) \mapsto x \] is clearly continuous

Consider \begin{align*} & \phi \circ \psi (x,f(x)) = \phi (x) = (x,f(x)) = Id_{G} \\ & \psi \circ \phi (x) = \psi (x, f(x)) = x = Id_{X}. \end{align*} Thus, $\psi $ is the continuous inverse of $\phi $ and hence $G$ is homeomorphic to $X$.