To show that the graph of a continuous map is a topological manifold, I need (among other things) to proof the following:
Let $U$ be an open subset of $\mathbb{R}^n$, and $f: U \to \mathbb{R}^k$ a continuous map. Then the map $\phi_f: U \to \mathbb{R}^{n+k}$ that associates $x \in U$ to $(x,f(x))$ is continuous.
I do not know how to prove it. I tried to show that the pre-image of every open subset of $\mathbb{R}^{n+k}$ is open in $U$ but I did not succeed. Also, I should not use the topological notion of product space since showing that the graph of a continuous map is a manifold appears before product spaces in the book I am reading.
Whenever $X$, $Y$ and $Z$ are three topological spaces and we equip $Y\times Z$ with the product topology, a function $\phi\colon X\to Y\times Z$ is continuous if and only if ${\rm pr}_Y\circ \phi\colon X\to Y$ and ${\rm pr}_Z\circ \phi\colon X\to Z$ are continuous. This is a trivial consequence of the definition of product topology. In other terms, if $\phi(x) = (\phi_1(x),\phi_2(x))$, then $\phi$ is continuous if and only if both $\phi_1$ and $\phi_2$ are continuous.
In your case, $X = U$, $Y = \Bbb R^n$, $Z = \Bbb R^k$, $\phi(x) = (x,f(x))$, $\phi_1$ is the inclusion $U\hookrightarrow \Bbb R^n$ and $\phi_2=f$.
Don't let $\Bbb R^n$ distract you.