Proof:

If $k = 0$ then $K$ is a point and $S^n - K$ is homeomorphic to $\mathbb{R}^n$ and the result is true in this case. The proof now proceeds by induction on $k$. Assume that the result has been proved for $0 \leq k \leq m-1$ and let $h:K\longrightarrow I^m$ be a homeomorphism. Define the halves $I^{+}$ and $I^{-}$ as

$$I^{+} = \{(x_1, x_2,\dots, x_m) \in I^m\;/\; x_n \geq 1/2\}, \quad I^{-} = \{(x_1, x_2,\dots, x_m) \in I^m\;/\; x_n \leq 1/2\}$$

and note that $I^{+}\cap I^{-}$ is homeomorphic to the cube $I^{m-1}$. We construct the sets $K^{+} = h^{-1}(I^{+})$ and $K^{-} = h^{-1}(I^{-})$, and use the Mayer Vietoris sequence to the following open cover of $S^n - (K^{+}\cap K^{-1})$:

$$\{S^n - K^{+},\; S^n - K^{-}\}.$$

Since $K^{+}\cap K^{-}$ is homeomorphic to $I^{m-1}$, by induction hypothesis the end terms of the portion

$$\begin{CD} \phantom{X}H_{j+1}(S^n-K^{+}\cap K^{-})@> >> H_j(S^n -... ...S^n-K^{-})@> >> \phantom{X}\\ @> q_j >> H_j(S^n - K^{+}\cap K^{-})\\ \end{CD}$$

are zero if $j > 0$ whereas the left most group is zero if $j = 0$. In any case $(\kappa^{\prime}, -\kappa^{\prime\prime})$ is injective. Assume that for some $j > 0$, $H_j(S^n - K) \neq 0$. We choose $\zeta \in H_j(S^n - K)$, $\zeta \neq 0$ and it follows $\kappa^{\prime}(\zeta) \neq 0$ or $\kappa^{\prime\prime}(\zeta) \neq 0$. Let us assume that $\kappa^{\prime}(\zeta) \neq 0$. Since $K^{+}$ is homeomorphic to $I^m$, the process can be repeated subdividing $K^{+}$ into two pieces whose intersection is homeomorphic to $I^{m-1}$. Thus we construct a nested sequence of subsets

$$K = K_1 \supseteq K_2 \supseteq K_3 \supseteq \ldots$$

such that for each $p$, the map $\kappa_p:H_j(S^n - K)\longrightarrow H_j(S^n - K_p)$ induced by inclusion, maps $\zeta$ to a non zero element. Composing with $f_p:H_j(S^n-K_p)\longrightarrow \varinjlim H_j(S^n - K_p)$ one checks that for $p, q \in \mathbb{N}$,

$$f_p\circ \kappa_p = f_q\circ \kappa_q,$$

thereby providing a map

$$f:H_j(S^n - K)\longrightarrow \varinjlim H_j(S^n - K_p).$$

Since the intersection $\bigcap K_i$ is homeomorphic to $I^{m-1}$, by induction hypothesis, $\varinjlim H_j(S^n - K_p) = \{0\}$. Hence $f_p(\kappa_p(\zeta)) = 0$ for every $p$ and hence by theorem (40.2) (ii), for some $q \in \mathbb{N}$, $\kappa_q(\zeta) = 0$ which is a contradiction.

Turning to the case $j = 0$, assume that rank of $H_0(S^n - K)$ is atleast two. If we select points $x$ and $y$ lying in distinct path components of $S^n - K$, the cycle $\zeta = x - y$ in $S^n - K$ is not a boundary. As before we construct a nested sequence of compact sets $\{K_p\}$ with $\kappa_p(\zeta) \neq 0$ for each $p \in \mathbb{N}$. But since $S^n - \bigcap K_p$ has only one path component, $\iota_p\circ \kappa_p(\zeta)$ is a boundary where $\iota_p$ is the map induced by the inclusion $S^n - K_p \longrightarrow S^n - \bigcap K_p$ whence

$$f_p(\kappa_p(\zeta)) = 0$$

by (41.3). This in turn forces $\kappa_p(\zeta) = 0$ by theorem (40.2) (ii) and we have a contradiction.