Natural manifold topology vs. product topology Let $M = \mathbb{R} \times \mathbb{R}$, where the first factor is endowed with the discrete topology and the second with the usual Euclidean topology and so $M$ has the resulting product topology. Show that $M$ admits a differentiable structure such that the natural manifold topology on $M$ is the product topology, hence in particular not second countable.

While trying to solve this I encountered some contradiction. I am quite sure my reasoning fails somewhere, but I am not too sure where. Here's my approach:
$${}$$
( I ) $\enspace$ The product topology is the topology such that the canonical projections are all continuous. Equivalently, the preimages of open sets under the canonical projections form a subbasis. In the above case of $\mathbb{R} \times \mathbb{R}$ this means that the open sets are of the form
$$ \{ x \} \times (a,b) \quad , \qquad x \in \mathbb{R},  \; a < b \quad . \tag{$\ast$}$$
and unions thereof. (Correct?)
( II ) $\enspace$ The natural manifold topology is the topology such that for a maximal atlas $A_{max} = \{ \, (\phi_i, U_i ) \; | \; i \in I \, \}$ the set $\{ U_i \}_{i \in I}$ forms a basis.
( III ) $\enspace$ I now define an atlas as $A = \{ \, (id_{\mathbb{R}^2}, V ) \; | \; V \in \tau_{\text{euclidean}} \, \}$, which obviously consists of all identity maps on open sets with respect to Euclidean topology. I now want to show that all $V \in \tau_{\text{euclidean}}$ are also elements of $\tau_{\text{product}}$.
( IV ) $\enspace$ In order to show said property, I just have to express an open ball (because every open set in Euclidean topology is the union of such balls) in terms of these "sliced intervals" in $(\ast)$. So for an open Ball with radius $r > 0$ and center $(m_1,m_2) \in \mathbb{R}^2$:
$$ B_r(m) \enspace = \enspace \bigcup_{x \in (m_1-r,m_1+r)} \{x\} \times (m_2 - \varepsilon_x, m_2 + \varepsilon_x) \quad , \qquad \varepsilon_x = r \cdot \cos \big[ \tfrac{\pi}{2r}(x-m_1) \big] $$
Which shows that the open sets $V \in \tau_{\text{euclidean}}$ are also $V \in \tau_{\text{product}}$ and by the convenient choice of coordinate charts being the identity, these charts $id_{\mathbb{R}^2} : V \longrightarrow V$ are homeomorphisms with respect to $\tau_{\text{product}}$. This, however, would mean that $\tau_{\text{euclidean}} = \tau_{\text{product}}$, contradicting that $M$ is not second-countable.
$${}$$
Where am I wrong? Which concept do I misunderstand?
 A: Your intuition is confusing you. $M$ is not a two-dimensional manifold, it is a $1$-dimensional manifold.
If $\tau_d,\tau_e$ are the discrete and Euclidean topologies on $\mathbb R$ then $(\mathbb R,\tau_e)$ is a $1$-dimensional manifold, practically by definition.
But $(\mathbb R,\tau_d)$ is a $0$-dimensional manifold. This partly depends on whether your definition allows $0$-manifolds, but all the same rules work if we treat $\mathbb R^0$ as the space with a single point. Then a topology is a $0$-dimensional manifold if and only if it is discrete.
Then the result:

If $M$ and $N$ are manifolds of dimension $m,n$ then $M\times N$ with the product topology is a manifold of dimension $m+n.$

still works if $m$ or $n$ or both are zero.
Your basis in the first step shows this. There can be no $2$-dimensional chart on $\{x\}\times(a,b),$ which is trivially homeomorphic to $(a,b),$ which is obviously $1$-dimensional topology.
One way to get a better intuition is that any path $\gamma:[0,1]\to X,$ where $X$ is a discrete topology, is a constant path.
In particular, the path connected components of your $M$ are thus of the form $\{x\}\times\mathbb R.$ But the Euclidean $2$ charts you propose would use open subsets which aren’t path connected. They might “look connected,” but that is because we imagine $\mathbb R\times\mathbb R$ with particular metric.
The reality of the discrete topology on $\mathbb R$ is hard to imagine, but it might be worth thinking instead of $M’=\mathbb N\times\mathbb R.$ It is intuitive that $M’$ is, topologically, a union of disjoint lines.
The same is true for $X\times\mathbb R$ for any discrete space $X.$ It is homeomorphic to the disjoint union of $|X|$ copies of the real line. But we can’t really picture it when $X$ is uncountable.
