Finding the de Rham cohomology of an open subset of $ \Bbb{R}^{n} $ minus a point. Here’s my question:

Let $ n \in \mathbb{N}_{\geq 2} $. Suppose that $ U \subseteq \Bbb{R}^{n} $ is an open set and that $ x \in U $. Then show that
  $$
{H_{\text{dR}}^{n - 1}}(U \setminus \{ x \}) \neq 0.
$$

My thoughts:

I was trying to use that
  $$
                S
\hookrightarrow U \setminus \{ x \}
\hookrightarrow \Bbb{R}^{n} \setminus \{ x \}
$$
  (where $ S $ is a small sphere in $ U $ centered at $ x $) should induce a homotopy equivalence, so
  $$
      {H_{\text{dR}}^{n - 1}}(U \setminus \{ x \})
\cong {H_{\text{dR}}^{n - 1}}(S)
\cong \Bbb{R}.
$$

I get the feeling that there’s something horribly wrong with this line of thought, so I was wanting some help. I’m okay with a direct answer, as long as you think the solution alone would be illuminating.
P.S.: This is my first post. Any advice on how I could make my question better would be appreciated as well.
 A: This actually can be done by some basic advanced calculus technique. In particular the Stokes Theorem. 
First by a translation and a scaling, assume that $x$ is the origin and the unit sphere $S$ is in $U$. Let 
$$\eta = \sum_{i=1}^n (-1)^{i-1} x^i dx^1 \wedge \cdots \wedge \hat{dx^i} \wedge \cdots dx^n,\ \ \ \ \omega = \frac{1}{r^n} \eta,$$
where $r = \sqrt{(x^1)^2 + \cdots + (x^n)^2}$ . Then one can check that $\omega$ is a closed form. One can also check that it is not exact: Indeed, $\omega = \eta$ when restricted on the unit sphere. So 
$$\int_S \omega = \int_S \eta = \int_{\partial B} \eta = \int_B d\eta = n Vol(B), $$
where $B$ is the unit ball. On the other hand, if $\omega$ is exact, $\int_S\omega$ is zero by Stokes theorem again. 
Thus $\omega$ is a closed but not exact $n-1$ form on $U\setminus\{0\}$. Thus $H^{n-1}_{dR} (U \setminus \{0\})$ is not zero. 
A: You have the right idea, but you're line of thought is not entirely correct. The map
$$S^{n-1}
\hookrightarrow U \setminus \{ x \}
\hookrightarrow \Bbb{R}^{n} \setminus \{ x \}$$ is indeed a homotopy equivalence, so the composition
$$H^{n-1}_{dR}(S^{n-1})
\rightarrow H^{n-1}_{dR}(U \setminus \{ x \})
\rightarrow H^{n-1}_{dR}(\Bbb{R}^{n} \setminus \{ x \})$$ is an isomorphism. Since $H^{n-1}_{dR}(S^{n-1})\neq0$ we get $H^{n-1}_{dR}(\Bbb{R}^{n})\neq0.$ (Otherwise the composition would not be injective.)
However, your claim $H^{n-1}_{dR}(U\backslash\{x\})\cong \mathbb{R}$ is not correct. An open annulus $U\subseteq\mathbb{R}^2$ serves as a counterexample, since it holds $H^1_{dR}(U\backslash\{x\})\cong\mathbb{R}\oplus\mathbb{R}$.
