Below $M^\circ$ and $\partial M$ denote the manifold-interior and manifold boundary of the topological manifold $M$ respectively.
Result (?): $f:\mathbb{R}^m\to\mathbb{R}^n$ be continuous and $M\subseteq \mathbb{R}^m$ be an $m$-manifold (with the subset topology). Then the graph $\Gamma(f)$ of $f$ is an $m$-manifold with $$\Gamma(f)^\circ = f(M^\circ) \ \ \ \ \text{ and }\ \ \ \ \partial\Gamma(f) = f(\partial M).$$
Remark: the result holds if $M$ is an open set, a closed ball or a closed box.
Proof: The projection $\phi:\Gamma(f)\to M$ onto the 'first' argument is a homeomorphism. Since $M^\circ$ is open so is $\phi^{-1}(M^\circ)$. Therefore the chart $(\phi^{-1}(M^\circ), \phi)$ is an interior chart for any $p\in \phi^{-1}(M^\circ)$. If one could show that the chart $(\phi^{-1}(\partial M),\phi)$ is an exterior chart for any $p\in\phi^{-1}(\partial M)$, the result would follow.
Does the result hold? If so, how would one finish the proof. If not, what is a counterexample?
Let $h:M\to N$ be a homeomorphism between topological spaces. If $M$ is a (topological) manifold (with or without boundary), then this means that every point $x\in M$ has a neighbourhood $U\subseteq M$ with a homeomorphism $u:U\to V$ where $V\subseteq \mathbb{R}^{k+}$ is open. Here by $\mathbb{R}^{k+}$ I mean the half subspace of $\mathbb{R}^k$.
And so $h(U)$ becomes an open neighbourhood of $h(x)$ homeomorphic to some open subset of $\mathbb{R}^{k+}$ through $u\circ k$, where $k: h(U)\to U$ is just restriction of $h^{-1}$.
The above construction shows how we transfer charts from $M$ to $N$.
And in that way $N$ is manifold. Moreover $h$ maps interior to interior and boundary to boundary. Because neighbourhoods around boundary points and interior points are not homeomorphic (which can be deduced by removing a point and observing how homotopy behaves). Furthermore, if we consider differential structure on $M$, then this construction turns $N$ into a differential manifold. With that $h$ becomes a diffeomorphism.
And since $\Gamma(f)$ is homeomorphic to $M$, then indeed it is a manifold. Indistinguishable from $M$.