Open Neighborhood of Closed Set Consider $A$ to be a closed (compact and convex if needed) subset of metric space $(\mathbb{R}^n, d)$, then for any open neighborhood $M$ of $A$ ($A \subseteq M$), there exists a $\delta>0$ such that $A_\delta \subseteq M$ where $A_\delta = \bigcup_{\mathbf{x} \in A} B_\delta (\mathbf{x})$.
The statement holds intuitively true for me. But I am having trouble rigorously prove the statement above.
I am aware that since $M$ is open and $A \subseteq M$, by definition of open set, I can say that $\forall~\mathbf{x} \in A$, there exists a $\delta(\mathbf{x})>0$ such that $B_{\delta(\mathbf{x})} \mathbf{x} \subseteq M$. But in my question, $\delta$ is constant with respect to $\mathbf{x} \in A$.
I have been struggling on this for a while now, any help or hints would be much appreciated.
 A: As in the comment you can see that compactness is needed.
If you assume that $A$ is compact, then you can consider the map $f:A\to \Bbb R$ given by $$ f(x)=d(x, \Bbb R^n\setminus M)$$ for every $x\in A$. Since $A$ is compact, $f$ is continuous and $A\subset M$ then $f$ admits a positive global minimum , that is there exists $\delta>0$ such that $B(x,\delta)\subset M$ for every $x\in A$.
A: As indicated in the comments, compactness of $A$ is needed.  Now, assuming that $A$ is compact, for each $x \in A$ you can find $\delta_x$ such that $B_{\delta_x}(x) \subseteq M$.  Now observe that $\{ B_{\delta_x / 2}(x) \mid x \in A \}$ covers $A$, so there exists a finite subcover $\{ B_{\delta_{x_i} / 2}(x_i) \mid i \in I \}$.  We now set $\delta := \min_{i\in I} (\delta_{x_i} / 2)$.
Now suppose that $x \in A$.  Then $x \in B_{\delta_{x_i / 2}}(x_i)$ for some $i \in I$.  Therefore,
$$B_\delta(x) \subseteq B_{d(x, x_i) + \delta}(x_i) \subseteq B_{\delta_{x_i}}(x_i) \subseteq M.$$
Here, we used that $d(x, x_i) < \delta_{x_i} / 2$, and also $\delta \le \delta_{x_i} / 2$, so $d(x, x_i) + \delta < \delta_{x_i}$.
