I don't really know where to begin here. Any help is appreciated, thank you.
"Let K be a compact set in $\mathbb R^n$. You may assume the Bolzano-Weierstrass theorem: every sequence in K has a subsequence that converges to some point of K. Prove that every continuous function from K to $\mathbb R^2$ is uniformly continuous."
Think of it as a game. For uniform continuity to hold, I first give you any $ε > 0$ and then you can always give me a $δ > 0$ such that, no matter what point $x \in K$ that I pick, you can show me that $Im_f(B(x,δ)) \subseteq B(f(x),ε)$.
So if uniform continuity does not hold, I have a winning strategy. How would it go? There must be some $ε > 0$ that I can give you that ensures I win. This means that no matter what $δ > 0$ you then give me, I can pick some point $x \in K$ such that $Im_f(B(x,δ)) \nsubseteq B(f(x),ε)$. It turns out that it is enough (*) to apply this to a sequence of small positive values for $δ$, say $δ_n = 2^{-n}$ and this strategy would give us a sequence of points $x_0,x_1,x_2, ...$ satisfying the final condition. But now we can use Bolzano-Weierstrass theorem and extract a subsequence that converges to some point $c \in K$. Also $Im_f(B(x_n,δ_n)) \nsubseteq B(f(c,ε))$ for any $n \in \mathbb{N}$. But $B(f(c,ε))$ is the same open ball (for the whole sequence), and there is always a point in $y_n \in B(x_n,δ_n)$ that does not map into it under $f$. But by continuity of $f$, as $n \to \infty$, we have $f(y_n) \to f(c)$ since $y_n \to c$ (intuitively $y_n$ is stuck in a ball with centre going to $c$ and radius going to $0$; but prove it!), and hence $f(y_n) \in B(f(c,ε))$ as $n \to \infty$, which contradicts the definition of $y_n$.
Therefore uniform continuity of $f$ on $K$ holds.