Let $A$ and $B$ be two compact subsets of $\mathbb{R}$. Let $f:A \times B \to \mathbb{R}$ be a continuous function on $A \times B$.
For each $a\in A$, define $B_a=\{b\in B:f(a,b)=0\}$, and suppose each $B_a$ is a singleton, so we may define a $g:A\to\Bbb R$ such that $g(a)$ is the unique element of $B_a$ for each $a\in A$. Is $g$ a continuous function of $a$?
I tried the following. But I have the feeling it's not correct.
Assume $g$ is not continuous at $a \in A$, then there exists a sequence $\{a_n\} \subset A$ converging to $a$ in $A$, for which the sequence $\{b_n\}=\{g(a_n)\} \subset B$ does not converges to $b=g(a)$. Since $B$ is compact, by the Bolzano-Weierstrass theorem, $\{b_n\}$ has a subsequence $\{b'_n\}$ converging to some $b'\ne b$ as $\{b_n\}$ does not converges to $b$. Let $\{a'_n\} \subset A$ be the subsequence of $\{a_n\}$ induces by $\{b'_n\}$. Since $\{a_n\}$ converges to $a$ then every subsequence of $\{a_n\}$ converges to $a$ and $\{a'_n\}$ converges to $a$. Since $g(a)$ contains a unique element $b$, then $b'=b$ which is a contraction since $\{b'_n\}$ is a subsequence of $\{b_n\}$. (The previous sentence seems suspicious).
The claim follows.