Proof:

(i) The sphere $S^n$ is compact and connected and $\mathbb{R}P^n$ is the continuous image of $S^n$ under the projection map $\eta$.

(ii) Let $\eta : S^{n}\longrightarrow \mathbb{R}P^n$ and $p : \mathbb{R}^{n+1}-\{0\} \longrightarrow (\mathbb{R}^{n+1}-\{0\})/\sim$ be the projection maps. We have a continuous map $\phi : \mathbb{R}^{n+1} - \{0\} \longrightarrow \mathbb{R}P^n$ given by the prescription

$$\phi(x) = \overline{x}, \quad x \in \mathbb{R}^{n+1} -\{0\},$$

where $\overline{x}$ is the equivalence class of $x$ in $S^{n+1}/\sim$. Denoting by $[x]$ the equivalence class of $x$ in $(\mathbb{R}^{n+1} - \{0\})/\sim$, the associated map $\overline{\phi} : (\mathbb{R}^{n+1}-\{0\})/\sim \;\;\longrightarrow \mathbb{R}P^n$ given by

$$\overline{\phi}([x]) = \overline{x}.$$

It is readily checked that $\overline{\phi}$ is bijective and $\overline{\phi}\circ \eta = \phi$. The universal property now gives us the continuity of $\overline{\phi}$. Consider now the map

$$\psi : S^n\longrightarrow (\mathbb{R}^{n+1}-\{0\})/\sim$$

given by $\psi({\bf x}) = [{\bf x}]$ which is evidently continuous map. There is a unique map

$$\overline{\psi}: \mathbb{R}P^n \longrightarrow (\mathbb{R}^{n+1}-\{0\})/\sim$$

such that $\overline{\psi}\circ \eta = \psi$. By the universal property of the quotient we see that $\overline{\psi}$ is continuous. It is left as an exercise to check that $\overline{\psi}$ and $\overline{\phi}$ are inverses of each other. Proof of (iii) is left as an exercise.

We shall see later that the spaces are Hausdorff as well. The space $\mathbb{R}P^1$ is a familiar space and the proof of the following result will be left for the reader.