Your statement of Theorem 4 is missing an assumption on $K$, such as being convex, or at least homeomorphic to such a set (convex, closed, bounded). Without such an assumption, rotation of a circle gives a counterexample. Also, I think that in Theorem 4 you want the normed space to be complete, i.e., a Banach space.
Theorem 3 is contained in Theorem 4, because on a compact set every continuous map is compact. Theorem 4 cannot be easily obtained from Theorem 3 (I think) because if we tried to simply replace $K$ with $\overline{f(K)}$ (which is compact), we can't apply Theorem 3 because $\overline{f(K)}$ is not known to be convex.
Both 3 and 4 were stated and proved by Schauder in his 1930 paper Der Fixpunktsatz in Funktionalraümen, which is in open access. Here is Theorem 3:
Satz I. Die stetige Funktionaloperation $F(x)$ bilde die konvexe, abgeschlossene und kompakte Menge $H$ auf sich selbst ab. Dann ist ein Fixpunkt $x_0$, vorhanden, d.h. es gilt $F(x_0)=x_0$.
And this is Theorem 4 (in slightly less general version: the image of $F$ is assumed compact instead of relatively compact; possibly because the latter concept wasn't in use).
Satz II. In einem "B"-Raume sei eine konvexe und abgeschlossene Menge $H$ gegeben. Die stetige Funktionaloperation $F(x)$ bilde $H$ auf sich selbst ab. Ferner sei die Menge $F(H)\subset H$ kompakt. Dann ist ein Fixpunkt vorhanden.
("B"-Raume is what is now called a Banach space.) So, it is correct to call both Theorem 3 and Theorem 4 "Schauder's fixed-point theorem".
And yes, Theorems 1 and 2 follow by specialization of Theorem 3 or 4 to finite dimensions.
Hint: Consider the (continuous!) function $g:X \to \Bbb R$ given by
$$
g(x) = d(x,f(x))
$$
Why must $g$ achieve its minimum?
To do this in a manner similar to the way you originally planned: for each $n \in \Bbb N$, define $U_n = \{x \in X: d(x,f(x)) > 1/n\}$. Take a finite subcover.
Best Answer
Let $X=\{-1,1\}$ and let $f(x)=-x$.