HINTS:
(c) $S^1$ is compact, and continuous functions preserve compactness.
(a) After you’ve done (c), you should know what kind of subset of $\Bbb R$ the continuous image of $S^1$ must be, unless it’s just a single point. It’s a very nice kind of set: it’s connected (why?), but if you remove almost any of its points, what’s left is not connected. Show that if $x$ is one of these so-called cut points whose removal disconnects $f[S^1]$, there must be at least two distinct points $p,q\in S^1$ such that $f(p)=f(q)=x$.
(b) If you do (a) using the hint above, this one will come almost for free.
$\mathbb{R}P^2$ is not homeomorphic to the upper hemisphere. In fact, it is the quotient of $S^2$ obtained by identifying all antipodal points $p,-p$.
Hence to see that $f$ determines $\tilde{f}$, it suffices to verify that $f(p) = f(-p)$. But this is obvious from the definition.
To prove that $\tilde{f}$ is an embedding, it suffices to show that it is injective (use the compactness of $\mathbb{R}P^2$). In other words, we have to show that $f(x,y,z) = f(x',y',z')$ implies $(x',y',z') = \pm (x,y,z)$. So let
$$(*) \phantom{xx} (x^2-y^2,xy,yz,zx) = ((x')^2-(y')^2,x'y',y'z',z'x') .$$
Write $c = x + iy \in \mathbb{C}$, $c' = x' + iy' \in \mathbb{C}$. Then $c^2 = x^2 - y^2 + i(2xy)$, $(c')^2 = (x')^2 - (y')^2 + i(2x'y')$. $(*)$ shows that $c^2 = (c')^2$, i.e. $c'= \epsilon c$ with $\epsilon = \pm 1$. In other words, $(x',y') = \epsilon (x,y)$.
Case 1. $(x,y) = (0,0)$. Then $z,z' \in \{ -1, 1\}$, hence $(x',y',z') = \pm (x,y,z)$.
Case 2. $(x,y) \ne (0,0)$. W.lo.g. let $x \ne 0$. Then $\epsilon x z' = x'z' = xz$, hence $z' = \epsilon z$. This shows $(x',y',z') = \epsilon (x,y,z)$.
Concerning your final question: The image $R = \tilde{f}(\mathbb{R}P^2)$ agrees with the image $f(S^2)$. And in fact $f(S^2) = f(S^2_+)$, where $S^2_+$ is the upper hemisphere of $S^2$. But note that $f \mid_{S^2_+}$ is no embedding because $f \mid_{S^2_+}$ still identifies antipodal points in the equator $E$ of $S^2$ which is contained in $S^2_+$.
Best Answer
$X$ compact does not imply that every homeomorphism of $X$ with itself has a fixed point. Consider $S^1$ where the homeomorphism is rotation by $\pi/2$. You need a better argument here.
This is incorrect. Hint: think about counting special types of points in $[0,1]$ and $(0,1)$.
Also incorrect. Think about compactness here.
What is the "complane"? If the question is "Is there a surjective continuous mapping from the complex plane to the non-zero reals?", then you should draw pictures of both of these sets and look at them- what property does one have which the other does not? Can you use this?