Basic topology question regarding open covers Let $(X,\tau)$ be a topological space. For a set $Y\subseteq X$ let $\text{int}Y$ denote the interior of $Y$ in $\tau$. Suppose we are given a collection of open sets $(V_{\alpha})_{\alpha\in A}$ in $\tau$ (where $A$ is some index set possibly infinite) such that $\text{int}Y\subseteq \bigcup_{\alpha\in A}V_{\alpha}$ but $Y\nsubseteq\bigcup_{\alpha\in A}V_{\alpha}$. 
Does there always exist some collection of open sets $(W_{\beta})_{\beta\in B}$ in $\tau$ such that $Y\subseteq \bigcup_{\beta\in B}W_{\beta}$ and $W_{\beta}\cap(\bigcup_{\alpha\in A}V_{\alpha})=V_{\beta}$ for all $\beta\in B$?  
added: $B\subseteq A$.
 A: The answer is no in general, that is such a collection $(W_{\beta})_{\beta\in B}$ need not exist. 
Let $X=\Bbb R^2$ be the plane, let $H$ be the horizontal strip $H=\{(x,y):0<y<2\}$, let $Q=\{(r,0):r\in\Bbb Q\}=\Bbb Q\times\{0\}$ where $\Bbb Q$ denotes the set of all rational numbers. Let $Y=H\cup Q$. Clearly $\mathrm{int}Y=H$. 
Let $A=\Bbb R$, i.e. we are going to use the reals as an index set. For each $t\in A$ let $V_t=B((t,1),1)=$ the open ball centered at $(t,1)$ and of radius $1$. 
Then $\mathrm{int}Y=H=\bigcup_{t\in A}V_t$ and $Y\nsubseteq\bigcup_{t\in A}V_t$. 
Next we want some index set $B\subseteq A$ and some $W_t$ open in $X$, for each $t\in B$, with $Y\subseteq \bigcup_{t\in B}W_t$ and $W_t\cap(\bigcup_{t'\in A}V_{t'})=V_t$ 
for all $t\in B$. That is, we want that 
$W_t\cap H=V_t$ for all $t\in B$. 
There is no way to obtain that. Given any rational $r$ there ought to be some $W_t$ with $(r,0)\in W_t$. But then there is some $\varepsilon>0$ such that the open ball $B((r,0),\varepsilon)\subseteq W_t$. But then 
$W_t\cap H\supseteq B((r,0),\varepsilon)\cap H$, and $B((r,0),\varepsilon)\cap H\nsubseteq V_t$ (the latter holds for every $t\in A$), so we cannot have 
$W_t\cap H=V_t$. 
