Linking topological spheres Is there a simple proof of the fact that: 

If $A\subset S^3$ is homeomorphic to $S^1$, then there is a circle $B$
  embedded into $S^3\setminus A$ that such that the circles $A$ and $B$
  are linked with the linking number $1$?

If $A$ is smoothly embedded, than it is easy. The problem is that the general topological embedding can be very complicated and I do not see a simple geometric explanation of this fact. I know that $H_1(S^3\setminus A)=\mathbb Z$ (Corollary 1.29 in Vick's Homology Theory) so the generator of the homology group should be a circle. 
Edit 1:
This is actually true as John Klein pointed out in his answer below, but I do not know the answer to the higher dimensional version of the problem stated below. Only a partial answer is given in comments below.
The question can be then generalized to higher dimensional spheres. 

If $A\subset S^n$ is homeomorphic to $S^k$, there should be a
  topological sphere in $S^n\setminus A$ of dimension $n-k-1$ that links
  $A$ with the linking number $1$.

Edit 2. A counterexample is given below in my answer.
 A: Let me add another answer. This answer stems from the fruitful discussion on MathOverflow.

Theorem.  There is a topological embedding $\iota:\mathbb{S}^1\to\mathbb{S}^5$ such that $\pi_3(\mathbb{S}^5\setminus\iota(\mathbb{S}^1))=0$.
Therefore, no
$3$-sphere can be linked with $\iota(\mathbb{S}^1)$.

Proof.
It is well known that there are $3$-dimensional integer homology spheres whose universal cover is $\mathbb{R}^3$. For example, there are particular constructions in [1,2,5] of hyperbolic integer homology spheres. Note that the universal cover of a hyperbolic $3$-manifold is the hyperbolic space that is homeomorphic to $\mathbb{R}^3$.
Other examples are listed in Homology sphere with $\mathbb{R}^3$ as the universal cover.
Let $\mathcal M$ be such an integer homology sphere. Since the universal cover of $\mathcal M$ is contractible, $\pi_3(\mathcal M)=0$. According to the celebrated theorem of Cannon and Edwards [3,4], the double suspension of an integer homology sphere is homeomorphic to a topological sphere. Let $h:S^2{\mathcal M}\to\mathbb{S}^5$ be such a homeomorphism.
$\mathcal M$ is a deformation retract of the complement of the vertices of the suspension $S\mathcal M$.
Therefore, $\mathcal M$ is also a deformation retract of the complement of the suspension of the vertices in $S^2\mathcal M$. Denote the suspension of the vertices by $X$, so $\mathcal M$ is a deformation retract of $S^2{\mathcal M}\setminus X$ and hence $\pi_3(S^2\mathcal{M}\setminus X)=0$.
$X$ is homeomorphic to $\mathbb{S}^1$. If $g:\mathbb{S}^1\to X$ is a homeomorphism, then $\iota=h\circ g:\mathbb{S}^1\to\mathbb{S}^5$ is a topological embedding and clearly $\pi_3(\mathbb{S}^5\setminus\iota(\mathbb{S}^1))=\pi_3(\mathbb{S}^5\setminus h(X))=\pi_3(S^2{\mathcal M}\setminus X)=0$.
The proof is complete. $\Box$
Edit. While the result is a straightforward consequence of known results, I decided to publish it as a short note that I dedicated to MathOverflow [6].
[1] J. Baldwin, J., S. Sivek,
Stein fillings and $SU(2)$ representations. Geom. Topol. 22 (2018), 4307-4380.
[2] J. Brock, N. M. Dunfield,
Injectivity radii of hyperbolic integer homology $3$-spheres.
Geom. Topol. 19 (2015),  497-523.
[3] J. W. Cannon, Shrinking cell-like decompositions of manifolds. Codimension three. Ann. of Math. 110 (1979), 83-112.
[4] R. D. Edwards, The topology of manifolds and cell-like maps. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 111–127, Acad. Sci. Fennica, Helsinki, 1980.
[5] J. Hom, T. Lidman,
A note on surgery obstructions and hyperbolic integer homology spheres.
Proc. Amer. Math. Soc. 146 (2018), 1363-1365.
[6] P. Hajłasz,  Linking topological spheres.  Atti Accad. Naz. Lincei Rend. Lincei Mat. Appl.  30 (2019), 907-909.
https://arxiv.org/abs/1906.01771
A: 
The question can be then generalized to higher dimensional spheres. If
  $A\subset S^n$ is homeomorphic to $S^k$, there should be a topological
  sphere in $S^n\setminus A$ of dimension $n-k-1$ that links $A$ with
  the linking number $1$.

In general the answer is in the negative. Here is an example. The double suspension of the Poincare homological sphere $M^3$ is homeomorphic to $\mathbb{S}^5$. This is a well known Cannon-Edwards theorem [1,2,3]. The suspension circle is an embedding of $\mathbb{S}^1$ to $\mathbb{S}^5$. Thus Poincare 3-sphere $M^3$ is a deformation retract of $\mathbb{S}^5\setminus\mathbb{S}^1$. 
Since $|\pi_1(M^3)|=120$ and the degree of a map $f:\mathbb{S}^3\to M^3$ is a multiple of $|\pi_1(M^3)|=120$, It follows that the linking number between the suspension circle $\mathbb{S}^1\subset\mathbb{S}^5$ and any embedding of $\mathbb{S}^3$ to $\mathbb{S}^5\setminus\mathbb{S}^1$ must be a multiple of $120$.
[1] https://en.m.wikipedia.org/wiki/Double_suspension_theorem
[2] J. W. Cannon, Shrinking cell-like decompositions of manifolds. Codimension three. Ann. of Math. 110 (1979), 83-112.
[3] R. D. Edwards,
The topology of manifolds and cell-like maps. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 111–127, Acad. Sci. Fennica, Helsinki, 1980. 
A: Let $C = S^3 \setminus A$. Alexander duality says that 
$$
H_1(C) \cong H^1(A) \cong \Bbb Z\,  .
$$
Let $\alpha: S^1 \to C$ be any map representing a generator of $H_1(C)$ (every first homology class is spherical by the Hurewicz theorem). By homotopical approximation we can assume $\alpha$ is a smooth embedding. We can also assume without loss in generality that the image of $\alpha$ misses the north pole of $S^3$. Identify the complement of the north pole with $\Bbb R^3$. Set $B= \alpha(S^1)$. Then $A\amalg B\subset \Bbb R^3$. The degree of the map 
$\require{AMScd}$
\begin{CD}
\ell: A \times B @>>>  S^2
\end{CD}
given by $(x,y) \mapsto (x - y)/|x - \alpha(y)|$ is the linking number of $A$ with $B$ by definition.  On the other hand, this map has degree one (after choosing appropriate homology generators). 
The point is that the pushforward of a generator 
\begin{CD}
H_1(S^1) @>\alpha_\ast >> H_1(C)
\end{CD}
coincides with the degree of $\ell$. 
How can we check this? Well, assuming $A$ has a nice regular neighborhood, we could redefine $C$ as the complement of that neighborhood. Then $\ell$ can be redefined as the degree of the map
$$
A \times C \to S^2
$$
again given by the same formula,
where we are assuming our new $C$ misses the north pole of $S^3$.
Alexander duality says that the induced slant product pairing 
$$
H_1(A) \otimes H_1(C) \to H_2(S^2) = \Bbb Z
$$
is non-singular, so the degree is $\pm 1$.
Even if $A$ fails to have a nice regular neighborhood, we can assume it misses the north pole $x$ and that $C:= S^n \setminus A$ misses another point $y$ of $S^n$. Identify $S^n \setminus x \cong \Bbb R^n \cong S^n \setminus y$. Then, similarly, we obtain a map
 $$
A\times C \to S^2
$$
and there a linking map $A\times B\to S^2$. With these changes, the argument proceeds as before.
