Show that $\mathbb{R}^n\setminus \{0\}$ is simply connected for $n\geq 3$

The path connectedness is easy. Given $x,\, y \in \mathbb{R}^n\setminus\{0\}$, the straight line segment $t \mapsto (1-t)\cdot x + t\cdot y$ connects $x$ and $y$ in $\mathbb{R}^n\setminus\{0\}$ unless $y$ is a negative multiple of $x$. If $y = c\cdot x$ for a $c < 0$, then you can compose a path of two straight line segments, one going a little away from $x$ to $x + \varepsilon\cdot e_i$ for a small enough $\varepsilon > 0$ and $i$ such that $x$ is not a multiple of $e_i$.

It could also be easily seen from the homeomorphism

$$F\colon S^{n-1} \times (0,\,\infty) \to \mathbb{R}^n\setminus\{0\};\quad (\xi,\,r) \mapsto r\cdot \xi.$$

(A proof that $F$ is indeed a homeomorphism may be required, or that may be considered obvious, depending on what was treated previously.)

Given a closed path $\gamma \colon t \mapsto (\xi(t),\, r(t))$ in $S^{n-1}\times (0,\,\infty)$, the map $H(s,t) = (\xi(t),\, r(t)^{1-s})$ provides a homotopy to a closed path in $S^{n-1}$.

If you already know that $S^d$ is simply connected for $d \geqslant 2$, you are done now.

Otherwise, to prove that fact, consider a closed path $\gamma\colon [0,\,1] \to S^{n-1}$. If $\gamma([0,\,1]) \neq S^{n-1}$, assume without loss of generality that the north pole $N = (0,\,\ldots,\,0,\,1)$ is not on the trace of $\gamma$. Stereographic projection from the north pole gives a homeomorphism $S^{n-1}\setminus \{N\} \to R^{n-1}$ (proof may be required), and that shows that $\gamma$ is null-homotopic in $S^{n-1}\setminus \{N\}$, hence a fortiori in $S^{n-1}$.

If $\gamma$ is surjective, you can partition it into parts that each omit one of two opposing caps of the sphere.

Consider for example the two caps $T = \{x \in S^{n-1}\colon x_n \geqslant \frac23\}$, $B = \{x \in S^{n-1}\colon x_n \leqslant -\frac23\}$ and the belt $E = \{x\in S^{n-1}\colon \lvert x_n\rvert \leqslant \frac13\}$.

By uniform continuity of $\gamma$, there is a $\varepsilon > 0$ such that $\gamma(t) \in B \land \lvert s-t\rvert < \varepsilon \Rightarrow \gamma(s) \notin E \cup T$, and similar for all other combinations of $B,\,E,\,T$.

Without loss of generality, suppose that $\gamma$ starts (and ends) in the south pole.

Set $t_0 = 0$. While $t_k < 1$, find the next partition point $t_{k+1}$ in the following way:

• If $\gamma(t_k) \in B$, let $t_{k+1} = \min\bigl(\{s > t_k \colon \gamma(s) \in E\} \cup \{1\}\bigr)$.
• If $\gamma(t_k) \in E$, let $t_{k+1} = \min \{s > t_k \colon \gamma(s) \in T\cup B\}$.
• If $\gamma(t_k) \in T$, let $t_{k+1} = \min \{s > t_k \colon \gamma(s) \in E\}$.

By the uniform continuity mentioned above, $t_m = 1$ for an $m \leqslant \frac{1}{\varepsilon}$.

Let $\gamma_k$ be the restriction of $\gamma$ to $[t_k,\,t_{k+1}]$.

If $\gamma(t_{k}) \in T$, then the composition $\gamma_{k-1}\gamma_k$ is a path connecting two points $a,\, b \in E$ in $S^{n-1}\setminus B$. The latter is homeomorphic to an open ball, so $\gamma_{k-1}\gamma_k$ is homotopic to a path connecting $a$ and $b$ in $E$.

Replacing all the $\gamma_{k-1}\gamma_k$ with homotopic paths in $E$, you obtain a path homotopic to $\gamma$ whose trace omits $T$, hence by the above, it, and therefore also $\gamma$ is null-homotopic in $S^{n-1}$.

The easy part of the Seifert-van Kampen theorem and which does not require an algebraic context (e.g. groups or groupoids) is that if $X$ is the union of a family $\mathcal U$ of path connected and simply connected open sets $U_i$ such that any pairwise intersection $U_i \cap U_j$ is path connected, and all containing a "base point" $a \in X$, then $X$ is simply connected. The key part of the proof is conveyed by the following diagram

in which $\alpha$ is a loop in $X$ at the base point $a$ subdivided so that each part $\alpha_i$ lies in a set say $U_i$ of $\mathcal U$. The start point $x_i$ of $\alpha_i$ shown as $\circ$ need not be at the base point. However $x_i$ lies in $U_{i-1} \cap U_i$ which is path connected; so there is a path $\gamma_i$ in $ U_{i-1} \cap U_i $ joining $x_i$ to the base point, and we set $\gamma_i$ to be constant for $i=0,n$. Now each path $\zeta_i= \gamma_i^{-1}\alpha_i\gamma_{i+1} $lies in $U_i$ which is simply connected, and so is homotopic in $U_i$ rel end points to a constant path $\beta_i$. Hence $\alpha$ is null homotopic, using the "horizontal" composition of these homotopies.

Note that even for this result we need a " horizontal composition of squares", and such an intuition is the start of a $2$-dimensional theory.

This type of argument was generalised by J.F. Adams (in unpublished lecture notes) to prove any map $S^r \to S^n$ is null homotopic for $r <n$, see 7.6.1 of Topology and Groupoids.

You can sledgehammer proof your problem by induction on the number of points that you remove and the Seifert-van Kampen theorem. Say $x_1,\ldots,x_m$ are the points that you remove. We will induct on $m$. When $m = 1$ this is easy because $\Bbb{R}^n$ minus a point is homotopy equivalent to $S^{n-1}$ that is simply connected for $n\geq 3$. For the general case of $x_1,\ldots,x_m$ points removed, I claim we can find a hyperplane $H$ that separates the first $x_1,\ldots,x_l$ points and last $x_{l+1},\ldots,x_m$ points. Let $U$ be the open cube minus contains the first $l$ points and $V$ the open cube minus the rest of the points. Then $\Bbb{R}^n - \{x_1,\ldots,x_m\}) = U \cup V$ and then Seifer-van Kampen says that

$$\pi_1(\Bbb{R}^n - \{x_1,\ldots,x_m\}) \cong \pi_1(U)\ast \pi_1(V)/\operatorname{some subgroup}.$$

We don't care about the subgroup at the moment because by induction $\pi_1(U) = \pi_1(V) = 0$ and so $\pi_1(\Bbb{R}^n - \{x_1,\ldots,x_m\}) = 0$. Done.

Edit: For those not convinced $S^{n-1}$ is simply connected for $n \geq 3$ here's a proof: Let $U$ be the upper hemisphere and $V$ the lower hemisphere, choose epsilon fattenings of these if you wish. For $n \geq 3$, $U \cap V$ is homotopy equivalent to the equator $S^{n-2}$ that is path - connected for $n \geq 3$. $U,V$ are homotopy equivalent to $D^{n-1}$ which is simply connected for $n \geq 3$. The hypotheses of the Seifert-van Kampen theorem are now satisfied and applying it to $U$ and $V$ shows $\pi_1(S^{n-1}) = 0$ for $n \geq 3$.