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For your first question, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "knot group"):

Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.

$\Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$\Big)$

For your third (and second) question, in generality, here's an argument that there is no such embedding. Any letter $S$ denotes a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:

(1) Up to ambient isotopy, there are only three embeddings of $S_0\sqcup S_1$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

$\Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.$\Big)$

(2) $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.

(3) An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.

$\Big($ Why? Here's one proof, which I found kind of fun: Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial M$ must lie outside both spheres of $\partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$\Big)$

(4) By process of elimination, $\partial M$ must be of the "avocado" type, so by (1) it's ambient isotopic to $A^\pm$, which by (2) is not ambient isotopic to the "side-by-side" arrangement $B$, which is what you wanted.

For your first question, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "knot group"):

Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.

$\Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$\Big)$

For your third (and second) question, in generality, here's an argument that there is no embedding from $M=S^{n-1}\times I$ to $\mathbb{R}^n$ with boundary ambient isotopic to the "side-by-side" embedding of
$S_0\sqcup S_1$. By any letter $S$ I denote a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:

(1) Up to ambient isotopy, there are only three embeddings of $S_0\sqcup S_1$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

$\Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.$\Big)$

(2) $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.

(3) An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.

$\Big($ Why? Here's one proof, which I found kind of fun: Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial M$ must lie outside both spheres of $\partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$\Big)$

(4) By process of elimination, $\partial M$ must be ambient isotopic to $A^\pm$, which by (2) is not ambient isotopic to the "side-by-side" embedding $B$, which is what you wanted.

For your first question, look at complements:

(One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.

Moral: considering complements is often useful in analyzing ambient isotopies, ambient homeomorphisms, etc. For example, this is a fundamental technique in knot theory (not an intentional pun when I first typed it): to any knot is an associated "knot group", defined to be the fundamental group of the compliment of the knot in $\mathbb{R}^3$.)

1. Up to ambient isotopy, there are only three embeddings of $S_0\sqcup S_0$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

(That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.)

2. $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.

3. An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.

Why? Here's one proof (this was fun): Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial(M))$ must lie outside both spheres of $\partial M$ (if it lies inside one, it can't touch the other). Let $x$ be the center of of the $S_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$, which must be a boundary point of $M$ (since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$). But it can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$ (this is now a Euclidean geometry argument which I'll stop rigorizing here). So $p$ is nowhere... contradiction.

4. By process of elimination, $\partial M$ must be of the "avocado" type, so by (1) it's ambient isotopic to $A^\pm$, which by (2) is not ambient isotopic to the "side-by-side" arrangement $B$, which is what you wanted.

For your first question, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "knot group"):

$\Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$\Big)$

(1) Up to ambient isotopy, there are only three embeddings of $S_0\sqcup S_1$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

$\Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.$\Big)$

(2) $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.

(3) An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.

$\Big($ Why? Here's one proof, which I found kind of fun: Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial M$ must lie outside both spheres of $\partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of  
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$\Big)$

(4) By process of elimination, $\partial M$ must be of the "avocado" type, so by (1) it's ambient isotopic to $A^\pm$, which by (2) is not ambient isotopic to the "side-by-side" arrangement $B$, which is what you wanted.

One trick for your first question is to look at compliments:

Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.

(One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.)

Moral of the story: considering compliments is often useful in analyzing ambient isotopies, ambient homeomorphisms, etc. For example, this is a fundamental technique in knot theory (not an intentional pun when I first typed it): to any knot is an associated "knot group", defined to be the fundamental group of the compliment of the knot in $\mathbb{R}^3$.

For your first question, look at complements:

Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $\mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $\mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=\mathbb{R}^3\setminus S$ to $V=\mathbb{R}^3\setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.

(One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.

Moral: considering complements is often useful in analyzing ambient isotopies, ambient homeomorphisms, etc. For example, this is a fundamental technique in knot theory (not an intentional pun when I first typed it): to any knot is an associated "knot group", defined to be the fundamental group of the compliment of the knot in $\mathbb{R}^3$.)

For your third (and second) question, in generality, here's an argument that there is no such embedding. Any letter $S$ denotes a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:

1. Up to ambient isotopy, there are only three embeddings of $S_0\sqcup S_0$ in $\mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

(That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.)

2. $A^\pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^\pm$ has one, $B$ has two.

3. An embedding from $M=S\times I$ to $R^n$ cannot have $\partial M = B$ up to ambient isotopy.

Why? Here's one proof (this was fun): Suppose $\partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=M\setminus \partial(M))$ must lie outside both spheres of $\partial M$ (if it lies inside one, it can't touch the other). Let $x$ be the center of of the $S_1$ component of $\partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$, which must be a boundary point of $M$ (since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$). But it can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$ (this is now a Euclidean geometry argument which I'll stop rigorizing here). So $p$ is nowhere... contradiction.

4. By process of elimination, $\partial M$ must be of the "avocado" type, so by (1) it's ambient isotopic to $A^\pm$, which by (2) is not ambient isotopic to the "side-by-side" arrangement $B$, which is what you wanted.