Open sets in a subspace topology Let $M$ be a subspace of a metrizable space $X$ and $U_i$, $i\in I$, are open sets in $M$. Prove that there exist open sets $V_i$, $i\in I$, in $X$ such that $U_i=V_i\cap M$ and for any finite $F\subseteq I$, if $\bigcap_{i\in F}U_i=\emptyset$ then $\bigcap_{i\in F}V_i=\emptyset$.
I am clueless about this problem.
Edit: No correct answer so far (two incorrect proofs were deleted). I think the idea is to write $V_i=\bigcup_{x\in U_i}B(x,r_{ix})$ for sufficiently small $r_{ix}>0$. But the difficulty lies in choosing $r_{ix}$ to capture the desired conclusion that $\bigcap_{i\in F}U_i=\emptyset\implies\bigcap_{i\in F}V_i=\emptyset$ for any finite $F\subseteq I$.
 A: For each $i \in I$ define $$V_i = \left\{ x \in X : d ( x , U_i ) < \tfrac{1}{2} d ( x , M \setminus U_i ) \right\}.$$  The sets $V_i$ are clearly open in $X$, and $V_i \cap M = U_i$.
Let $F \subseteq I$ be finite such that $\bigcap_{i \in F} U_i = \emptyset$.  Suppose that $x \in \bigcap_{i \in F} V_i$.  We can recursively construct sequences $\langle i_k \rangle_{k=0}^\infty$ and $\langle y_k \rangle_{k=0}^\infty$ so that


*

*$i_k \in F$;

*$y_k \in U_{i_k}$;

*$d ( x , y_k ) < \frac{1}{2} d ( x , M \setminus U_{i_k} )$; and

*$y_k \notin U_{i_{k+1}}$


As $d ( x , y_{k+1} ) < \frac{1}{2} d ( x , M \setminus U_{i_{k+1}} ) \leq \frac{1}{2} d ( x , y_k )$ it is easy to see that $\lim_{k \rightarrow \infty} d ( x , y_k ) = 0$.  As $F$ is finite there is an $i \in F$ and a strictly increasing sequence $\langle k_j \rangle_{j=0}^\infty$ such that $i_{{k_j}+1} = i$ for all $j$.  As $y_{k_j} \in M \setminus U_{{k_j}+1} = M \setminus U_i$ for all $j$ it follows that $d ( x , M \setminus U_i ) \leq \lim_{j \rightarrow \infty} d ( x , y_{k_j} ) = 0$, contradicting that $x \in V_i$!
