If $\dim_{\mathbb{R}}S<n$, then

$$H^{n}(X,\mathbb{C})=H^{n}(T_{r}(S),\mathbb{C})=H^{n}(S,\mathbb{C})=\{0\}$$

for any $r>0$, where $T_{r}(S)=\{p\in X|dist_{g_{0}}(p,S)\leq r\}$. Let $r_{0}>r_{1}>\sigma$ such that $T_{r_{1}}(S)\subset B_{g_{0}}(x_{0},r_{0})$, and $F_{r_{0},k}(T_{r_{1}}(S))\supset B_{\tilde{g}_{k}}(x_{k},\sigma)$ for $k\gg 1$. Then the inclusion maps induce homeomorphisms on cohomology groups

$$H^{n}(M,\mathbb{C})\longrightarrow H^{n}(F_{r_{0},k}(T_{r_{1}}(S)),\mathbb{C})\longrightarrow H^{n}(B_{\tilde{g}_{k}}(x_{k},\sigma),\mathbb{C}),$$

$${\rm and}\ \ \ [\tilde{\Omega}_{k}]\mapsto[\tilde{\Omega}_{k}|_{F_{r_{0},k}(T_{r_{1}}(S))}]\mapsto[\tilde{\Omega}_{k}|_{B_{\tilde{g}_{k}}(x_{k},\sigma)}]\neq 0.$$

Thus $[\tilde{\Omega}_{k}|_{F_{r_{0},k}(T_{r_{1}}(S))}]\neq 0$ in $H^{n}(F_{r_{0},k}(T_{r_{1}}(S)),\mathbb{C})$, which contradicts to

$$H^{n}(F_{r_{0},k}(T_{r_{1}}(S)),\mathbb{C})\cong H^{n}(T_{r_{1}}(S),\mathbb{C})=\{0\}.$$

∎