What is the smallest Euclidean space in which one can embed a given curved space?

Like you say:

• The plane $\Bbb R^2$ can be embedded in $\Bbb R^2$ but not $\Bbb R$.
• The sphere $\Bbb S^2$ can be embedded in $\Bbb R^3$ but not $\Bbb R^2$.

In particular, the answer depends not only on the dimension of the given curved space (which here I'll take to mean manifold), but the space itself.

We also have:

• Real projective space $\Bbb {RP}^2$ and the Klein bottle $K$ can both be embedded in $\Bbb R^4$ but not $\Bbb R^3$.

In fact, $4$ dimensions is sufficient for all surfaces ($2$-manifolds); see below.

To be more precise the answer depends on whether one means a smooth embedding or a Riemannian embedding, that is, whether one has a metric on the manifold one wants to line up with the Euclidean metric of the ambient space. (The results above refer to smooth embeddings.)

Smooth manifolds

The (Strong) Whitney Embedding Theorem gives a universal upper bound: A smooth $m$-manifold can be embedded in $\Bbb R^{2m}$.

Under some common conditions, we can improve this bound some. Wall ($m = 3$) and Haefliger-Hirsch ($m > 3$) showed that if a smooth $m$-manifold is compact and $m$ is not a power of $2$, then it can be embedded in $\Bbb R^{2m - 1}$. (We can even improve on this, but there is no known closed-form formula for the maximum dimension required as a function of $m$.) This restriction re powers of $2$ is necessary, as $\Bbb {RP}^{2^k}$ cannot be embedded in $\Bbb R^{2 \cdot 2^k - 1} = \Bbb R^{2^{k + 1} - 1}$.

The Weak Whitney Embedding Theorem says that we can do slightly better if we allow our manifold to be immersed, rather than embedded (roughly speaking, this means we allow self-intersections that aren't pathological): A (topological, so not necessarily smooth) $m$-manifold can be embedded in $\Bbb R^{2 m - 1}$. (In particular, there is a smooth immersion of $\Bbb {RP}^2$ into $\Bbb R^3$, despite that there is no such embedding.

Riemannian manifolds

If one wants to embed a (smooth) Riemannian manifold $(M, g)$ in Euclidean space $(\Bbb R^N, \bar{g})$, so that the metric $g$ is the metric determined on $M$ by the embedding, one generally requires higher-dimensional Euclidean spaces than Whitney's Embedding Theorem prescribes: The Nash Embedding Theorem (well, one theorem by this name) says that a Riemannian $m$-manifold can be embedded in $\Bbb R^{m (3m + 11) / 2}$ if the manifold is compact and $\Bbb R^{m (m + 1) (3m + 1) / 2}$ if not.

NB that the dimension required can depend on $g$ and not just on $M$. For example, the torus $\Bbb T^2$ can be embedded into $\Bbb R^3$, and any such embedding determines a metric on it. No such metric is flat, however, so the flat torus cannot be embedded into $\Bbb R^3$; it can, though, be embedded in $\Bbb R^4$.