When $d=1$ it is of course possible to embed large chunks of $S^d$ isometrically in ${\mathbb R}^n$.
When $d\geq2$ it is not possible to embed even tiny pieces of $S^d$ isometrically in ${\mathbb R}^n$.
Proof. Take any three points $x_1$, $x_2$, $x_3\in S^d$ forming a small equilateral triangle in the metric of $S^d$. This then is an "ordinary" spherical triangle $\triangle$ with side length $s>0$ on the $2$-sphere $S^d\cap\langle x_1,x_2,x_3\rangle$. A map $f$ of the required kind will map the $x_i$ to the vertices $y_i$ of an equilateral triangle $\triangle'$ embedded in ${\mathbb R}^n$, and having side length $s$ as well. The distance from a vertex $y_i$ to the midpoint of the opposite side in $\triangle'$ amounts to ${\sqrt{3}\over2}s$, which is certainly different from the corresponding distance in $\triangle$.
Note that by definition, $ d(x, y) \ge \| x-y\|_E$ for all $x, y\in M$, so
$$ \frac{d(x, y)}{\|x-y\|_E} \ge 1, \ \ \ \forall x, y\in M, x\neq y.$$
So it suffices to show that
$$\tag{1}\limsup_{y\to x} \frac{d(x,y)}{\|x-y\|_E} \le 1$$
We assume that $M$ has dimension $k <n$. Using a rotation and translation, we assume that $x = 0\in M$ and $T_xM = \mathbb R^k \times \{0\} \subset \mathbb R^k \times \mathbb R^{n-k}$. Since $M$ is embedded in $\mathbb R^n$, there is $R>0$ and an open set $U$ with $0\in U\subset \mathbb R^k$ so that
$$ M\cap B_x(R) = \{ (z, u(z)) : z\in U\},$$
here $u: U \to \mathbb R^{n-k}$. Now let $y\in M\cap B_x(R)$. So $y = (z, u(z))$ for some $z$ and
$$ \| x-y\|_E = \sqrt{\|z\|^2 + \| u(z)\|^2}\ge \|z\|.$$
Note that we have $u(0) = \nabla u(0) = 0$. On the other hand, let $\gamma(t) = (tz, u(tz))$. Then this is a curve in $M\cap B_x(R)$ joining $x$ to $y$. Thus
\begin{align*}
d(x, y) &\le \int_0^1 \| \gamma'(t)\|_g dt\\
& = \int_0^1 \sqrt{\|z\|^2 + |\langle z, \nabla u (tz)\rangle|^2}\, dt
\end{align*}
where $\nabla u (tz)$ is the gradient of $u$ at the point $tz$ and $\langle \cdot, \cdot \rangle$ is the dot product on $\mathbb R^k$ (and we also used that $\gamma'(t) = (z, \langle z, \nabla u (tz)\rangle)$). Thus we have
$$ \frac{d(x, y) }{\|x-y\|_E} \le \frac{1}{\|z\|} \int_0^1 \sqrt{\|z\|^2 + |\langle z, \nabla u (tz)\rangle|^2} \, dt = \int_0^1 \sqrt{1 + |\langle \nu, u(tz)\rangle|^2} \, dt ,$$
where $\nu = z/\|z\|$. Since $\nabla u \to 0$ as $y\to x$, we conclude that
$$ \int_0^1 \sqrt{1 + |\langle \nu, u(tz)\rangle|^2} \, dt \to 1.$$
Thus (1) is shown.
Best Answer
The usual 2-sphere exists naturally in $\mathbb R^3$, and in general the usual definition of $S^n$ is as a particular subset of $\mathbb R^{n+1}$ with the induced metric. In that case, the identity map is a locally metric-preserving embedding into $\mathbb R^2$, but it doesn't preserve the global distance. To wit, two diametrically opposed points have distance $2$ in $\mathbb R^3$ but distance $\pi$ along geodesics in the sphere itself.
Thus, the natural embedding works as an isometry when we view the two spaces as Riemannian manifolds, but not when we consider them directly as metric spaces. It appears that both kinds of maps can be called "isometric embeddings", but nonetheless they are different concepts.