The proof of this is based on the idea given on page 21 here. It's about finding the fundamental groups of the sphere and the Projective Plane without using Van Kampen's theorem. It still seems like a weak form of it, though.
Let $\nu=(0,\dots,1),\xi=-\nu\in S^n$. Also let $I=[0,1]$.
Lemma 1. Let $\gamma$ be a loop in $S^n$ starting at $\xi$ such that $\gamma$ doesn't pass through $\nu$. Then $\gamma$ is nullhomotopic.
Proof. $\im\gamma\subseteq S^2-\nu\cong \bR^n$. Since $\bR^n$ is simply connected, $\gamma$ is nullhomotopic.
Lemma 2. Let $\gamma$ a loop in $S^n$ starting at $\xi$ which passes through $\nu$. Then $\gamma$ is homotopic to a loop in $S^n-\nu$.
Proof. Let $U$ be a small open ball around $\nu$ and $U'$ a bigger ball, such that $\overline{U}\subseteq U'$. Since $\gamma^{-1}(v)$ is closed in $I$ then it's compact. Now, $\gamma^{-1}(U)$ is a union of open, disjoint intervals which cover $\gamma^{-1}(\nu)$; take a finite number of them which still cover $\gamma^{-1}(\nu)$, say $\{I_k\}_{k=1}^n$ such that $\overline{I_k}=[a_k,b_k]$ for each $k$, and note that $\gamma(I_k)\subseteq \gamma(\gamma^{-1}(U))\subseteq U$. Also $\gamma(\overline{I_k})\subseteq U'$. Since $U'$ is simply connected, then, for each $k$, the curve $\gamma|_{\overline{I_k}}$ is homotopic to any curve $\gamma_k$ with the same endpoints not passing through $\xi$.
Then
$$[\gamma]=[\gamma|_{[0,a_1]}*\gamma_1*\gamma|_{[b_1,a_2]}*\cdots*\gamma_n*\gamma|_{[b_n,1]}]$$
where
$$\gamma|_{[0,a_1]}*\gamma_1*\gamma|_{[b_1,a_2]}*\cdots*\gamma_n*\gamma|_{[b_n,1]}$$
is a loop starting at $\xi$, and not passing through $\nu$.
Corollary. $S^n$ is simply connected.
Proof. Every loop on $S^n$ starting at $\xi$ is homotopic to a loop not passing through $\nu$, which is nullhomotopic.
Proposition. $\pi_1(\bR\bP^n,\overline{x})\cong \bZ/2\bZ$ for any $\overline{x}\in\bR\bP^n$.
Proof. Let $\overline{x}=\{x,-x\}$ with $x\in S^n$. Let $\gamma$ be a loop in $\bR\bP^n$ starting at $\overline{x}$. Then it lifts to a curve $\tilde{\gamma}$ in $S^n$ starting at $x$. If $\tilde\gamma$ is itself a loop, then $\tilde\gamma$ is nullhomotopic and then $\gamma$ is also nullhomotopic. Then, let's suppose $\tilde\gamma$ is not a loop; so that $\tilde\gamma(1)=-x$ and $\gamma$ is not nullhomotopic. Let $\delta$ be another loop at $\overline{x}$ and $\tilde\delta$ its lift at $-x$. Then $\tilde\gamma*\tilde\delta$ is a loop, and also lifts $\gamma*\delta$ at $x$. So $\tilde\gamma*\tilde\delta$, and $\gamma*\delta$ are both nullhomotopic. So every element of $\pi_1(\bR\bP^n,\overline{x})$ is an inverse of $[\gamma]$, thus $\pi_1(\bR\bP^n,\overline{x})$ has at most two elements. Since $\gamma$ is not nullhomotopic, $\pi_1(\bR\bP^n,\overline{x})$ has exactly two elements. So $\pi_1(\bR\bP^n,\overline{x})\cong \bZ/2\bZ$.
Note. We can also use an argument about the number of sheets of the covering space $S^2$: The cardinal of the fiber of $\overline{x}$ is in a $1-1$ correspondence to the cardinal of $\pi_1(\bR\bP^n,\overline{x})$. Since the former has two elements, the later too. I like the proof above more, though.
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