$X_1$, $X_2$ are closed and open connected components of the сompact $X = \sin (1/x)$. ( It is compact $X \subseteq \mathbb{R}^2$ - the union of a vertical segment $X_1 = \{(0,y) : −1 \leq y \leq 1\}$ and graph $X_2$ of function $y = \sin(1/x)$, $0 < x \leq 1$)
Prove that if we have a continuous function $f:X\rightarrow X$, then exist point $x \in X_2$: if $f(x)\in X_1$, then $f(X) \subseteq X_1$.
Please, help me understand how i can do it. And please excuse my English.
Note that $X$ (the topologist's sine curve) is connected, so $X_1$ and $X_2$ are not connected components of $X$. However, $X$ is not path connected, and $X_1, X_2$ are the path components of $X$. See Topologist Sine Curve, connected but not path connected. It is moreover easy to verify that $\overline{X}_2 = X$ and $\overline{X}_1 = X_1$.
Given a continuous $f : X \to X$, the images $f(X_i)$ are path connected and must therefore be contained either in $X_1$ or in $X_2$.
Under the assumption in your question we infer that $f(X_2) \subset X_1$. But then continuity implies $f(X) = f(\overline{X}_2) \subset \overline{f(X_2)} \subset \overline{X}_1 = X_1$.