Let $Y$ be locally connected and $A \subset Y$ arbitrary. Let $C$ be a component of $A$. Prove: a. $Int(C)= C \cap Int(A)$ b. $Fr(C) \subset Fr(A)$ c. If $A$ is closed, then $Fr(C)=C \cap Fr(A)$
My atempt:
b. For a. $$Fr(C)=\overline{C} - Int(C)= C \cap (C \cap Int(A))^c = C \cap (C^c \cup Int(A)^c)= C - Int(A) \subset A - Int(A) \subset \overline{A} - Int(A)= Fr(A)$$
Help me a. and c. Please.
As Henno Brandsma commented, components of open sets in a locally connected space $Y$ are open in $Y$. See for example About locally path-connected spaces.
Proof of a. : $\text{int}(C) \subset C \cap \text{int}(A)$ is trivial because $C \subset A$. Now let $x \in C \cap \text{int}(A)$ and let $D$ be the component of $x$ in $\text{int}(A)$. Clearly $D \subset C$. Since $\text{int}(A)$ is open in $Y$, also $D$ is open in $Y$. Hence $x \in D \subset \text{int}(C)$.
Proof of c. All components of a space $Z$ are closed in $Z$. Since $A$ is closed in $Y$ and $C$ is closed in $A$, we see that $C$ is closed in $Y$. Therefore $$\text{Fr}(C) = C \setminus \text{int}(C) = C \setminus (C \cap \text{int}(A))= C \cap (C \cap \text{int}(A))^c = C \cap (C^c \cup \text{int}(A)^c) \\ = C \cap (Y \setminus \text{int}(A)) = C \cap (\overline A \setminus \text{int}(A)) = C \cap \text{Fr}(A) .$$ Note that it would be sufficient to require that $C$ is closed in $Y$.