In our lecture notes, we have the following lemma which is afterwards used to prove the classification of closed 1-manifolds (Compact Hausdorff spaces every point of which has a neighbourhood homeomorphic to an open subset of $R^n$). The lemma has the following statement:
Let $X$ be a topological space and $U,V$ be open subsets of $X$. Consider two homeomorphisms $$\phi\colon U\rightarrow (0,1) \\ \psi\colon V\rightarrow (0,1) $$ Suppose that $U\cap V \not = \varnothing $ and let $K$ be a path component of $U\cap V$. We have $$ \phi(K)=(a,b)\subseteq (0,1) \\ \psi(K)=(c,d)\subseteq (0,1).$$ Then ($a=0$ or $b=1$) and ($c=0$ or $d=1$).
The proof available to me I don't quite understand. Moreover I don't understand why the image of $K$ under these homeomorphisms needs to necessarily be an open subinterval of $(0,1)$ as in the statement of the proof. I would like to ask for some guidance on this or perhaps some reference where I can find a good proof of this fact.
For your second question, the set $U \cap V$ is open in $U$. It follows that $\phi(U \cap V)$ is open in $\phi(U)=(0,1)$. The set $\phi(K)$ is a path component of $\phi(U \cap V)$. Every path component of an open subset of $(0,1)$ is a subinterval, which is proved by application of the intermediate value theorem. It follows that $\phi(K)$ is a subinterval.
For your first question, I suspect something is wrong. For a counterexample, having specified $U$ and $\phi$, we can then take $V = \phi^{-1}(1/3,2/3)$ and $\psi = \phi \mid V$. It follows that $V \subset U$ so $K = V \cap U = V$, hence $\phi(K) = \phi(V) = (1/3,2/3)$. But $a=1/3 \ne 0$ and $b=2/3 \ne 1$. Perhaps there is a missing hypothesis to the lemma.