Determining the dimension of manifold Spivak in Calculus on Manifolds states that a subset $M$ of $\mathbb{R}^n$ is a $k$ dimensional manifold $M$ if:
For every point $x \in M$ there exists open $U$ containing $x$ and open $V \subset \mathbb{R}^n$ and a diffeomorphism $h: U\rightarrow V$ such that $h(U \cap M) = V \cap (\mathbb{R}^k \times \{0\})$.
When I first saw this definition, I immediately wondered if a $k$ dimensional manifold was also a $k+1$ dimensional manifold. Upon some thought, it did not appear so.
For example, clearly, the xy plane in $\mathbb{R}^3$ is a 2-dimensional manifold, by taking $h$ as the identity. But intuitively we can't construct a differentiable function that takes $U \cap M$ to $V \cap \mathbb{R}^3$, since $M$ is "flat" and $V \cap \mathbb{R}^3$ (WLOG) is an open ball, so the xy plane is not a 3-dimensional manifold.
Question 1:
Is it true that if $M$ is a $k$ dimensional manifold, it must not be a $k'$ dimensional manifold for $k \neq k'$?
Question 2:
If the answer to Question 1 is True, then how does one see this easily in the general case?
 A: What you think is correct. If a manifold is of dimension $k$, then for any other $k' \neq k$, the manifold cannot be of dimension $k$. The answer lies in the fact that for any $n_1$ and $n_2$ (natural numbers), the topological spaces $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ are not homeomorphic.
Since while establishing the dimension of a manifold, we essentially use homeomorphisms of open sets of the manifold with open sets in $\mathbb{R}^n$, for a fixed open set in the manifold $M$ we cannot have two distinct $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ for which homeomorphisms are possible. Otherwise, it would mean two open sets in distinct $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ are homeomorphic, which is not possible. If you wonder why such a thing is not possible, just consider the case when the two open sets in $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ are connected. Then, they are homeomorphic (individually) to $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ respectively. Then, using composition, you must conclude that the two spaces $\mathbb{R}^{n_1}$ and $\mathbb{R}^{n_2}$ are homeomorphic. When the open sets in consideration are not connected, pick a point (preferably the image of the same point in the manifold) in the two sets and construct open balls around them. Then, the restriction of homomorphism (from the manifold to these sets) is again a homomorphism, and now you can apply the previous case.
In fact, that is why we say the dimension of a manifold and not a dimension.
PS: As you have mentioned the intuition of the $XY$-plane being "flat", you can think of the dimension of a manifold as the "number" of (linearly independent) axes you have at a given point. Now, if a particular point has two neighbourhoods that give a different number of axes, how many linearly independent axes do you have at that point? This is where the problem occurs if a manifold has more than one dimension at a point!
A: The magic words are "invariance of domain".
