Problem: Let $F_1$ and $F_2$ are closed set in a metric space $X$. Let $A = F_1 \cup F_2$. and $f:A \rightarrow Y$ is a defined function on $A$. Show that if $f|_{F_1}$ and $f|_{F_2}$ are continuous function so $f$ continuous on $A$.
My opinion: If $F_1$, $F_2$ are closed set $X$, so $A$ is closed set in $X$. We have $f|_{F_1}$ is a continuous function, hence $\exists (x_n)_n \subset F_1$ such that $(x_n)_n \rightarrow x_0 \in F_1$ then $f(x_n) \rightarrow f(x_0)$. $x_0 \in F_1 \Rightarrow x_0 \in A$ and thus $f$ continuous on $A$.
My question: Is this problem provide redundant information or there are some trap on this
To prove that $f$ is continuous on $A$ you have to start with an arbitrary sequence $\{x_n\}$ in $A$ converging to a point $x$ in $A$ and show that $f(x_n) \to f(x)$. So your argument is not valid. Here is a proof: since $x \in A$ either $x \in F_1$ or $x \in F_2$. Suppose $x \in F_1$. Split the sequence $\{x_n\}$ into two parts, one coming from $F_1$ and one from $F_2$. If there are infinitely many $n$ with $x_n \in F_2$ then $x =\lim x_n \in F_2$ (also) and you can use continuity of $f|F_2$ . Otherwise you can use continuity of $f|F_1$. Since we have convergence for both subsequences, the whole sequence $f(x_n) \to f(x)$. Similar argument when $x \in F_2$. Simpler argument: just consider $f:A \to Y$ and prove that the inverse image any closed set in $Y$ is closed in $A$. This is quite easy. [ Inverse image of $C$ under $f|F_i$ is $f^{-1}(C)\cap F_i$. Use the identity $[f^{-1}(C)\cap F_1] \cup [f^{-1}(C)\cap F_2]=f^{-1}(C)\cap A$.