This question was previously closed as a duplicate of this question, which asserts that if $f : X \to Y$ is a homeomorphism and that if $C$ is a connected component of $X$, then $f(C)$ is a connected component of $Y$, and $C$ is homeomorphic to $f(C)$. Per that linked question, if $f : (0,1)\cup (1,2) \to \{a,b\}$, then it must be that either $f((0,1)) = \{a\}$, or $f((0,1)) = \{b\}$. In either case, the restriction of $f$ to $(0,1)$ is a homeomorphism from an uncountable set to a singleton. By a cardinality argument, this can't happen. — Xander Henderson ♦ 2 mins ago

If the problem is understanding why homeomorphisms induce homeomorphisms of connected components, then this question is a duplicate of the linked question. If the problem is understanding why an interval is not homeomorphic to a singleton set, then a straight-forward cardinality argument resolves the issue (the cardinality of a singleton is 1, which is quite a lot smaller than the cardinality of an interval, which rules out the possibility of a bijection). In either case, it is not at all clear what the actual obstruction is, and so I don't believe that this question is suitable for Math SE. — Xander Henderson ♦ 20 secs ago