Can two closed disjoint subsets be at zero distance from each other? If $A,B$ are two closed subsets of a metric space $(M,d)$ such that $A\cap B=\emptyset$ and $B$ is sequentially compact then $d(A,B)>0$ where $d(A,B)=\inf\{d(a,b): a\in A, b\in B\}$
Do we need the sequential compactness of one of the sets to have $d(A,B)>0$?
If we do then would adding "$B$ is bounded" instead of the sequentiall compactness be sufficient?
NB: I think I have a counter example for my original question: $A=\{(x,y)\in \Bbb R^2 : x<0, y\ge-1/x\}$ and $B=\{(x,y)\in\Bbb R^2: x>0, y\ge 1/x\}$
 A: In the original set-up ($A$ and $B$ are disjoint closed subsets with $B$ sequentially compact), it must be the case that $d(A,B) > 0$, provided that you also assume both sets are non-empty.  One can prove so by contradiction.  
Suppose for contradiction that $d(A,B) = 0$.  By definition of $d(A,B)$, this would mean that for every $n \in \mathbb{N}$, we could find points $a_n \in A$ and $b_n \in B$ with $d(a_n, b_n) < \frac{1}{n}$.  By sequential compactness of $B$, we could extract a subsequence $\{ b_{n_k} \}$ so that $\{ b_{n_k} \}$ converges to a point $b$ in $B$.  It would follow from the triangle inequality that the sequence  $\{ a_{n_k} \}$ also converges to $b$.  Thus, $b$ would be a limit point of $A$ and would therefore be contained in $A$ by the assumption that $A$ is closed.  This contradicts the assumption that $A$ and $B$ are disjoint.
The hypothesis of sequential compactness is indeed necessary.
Your proposed counterexample to the original statement actually shows this, because your proposed set $B$ is not sequentially compact.  (To see that $B$ is not sequentially compact, consider the sequence $b_n = (\frac{1}{n}, n + \frac{1}{n})$.  There is no subsequence of $\{b_n \}$ that converges to a point in $B$, because the first coordinates of the points $b_n$ converge to $0$.  However, $x > 0$ for all points $(x,y) \in B$.  A similar argument shows that $A$ is not sequentially compact either.)
For another counterexample in the case that neither set is sequentially compact, let $M = \mathbb{Q}$ and let $d$ be the Euclidean metric inherited from $\mathbb{R}$, i.e. $d(x,y) = |x-y|$.  One can take $A = \{ x \in \mathbb{Q}:  0 \leq x \leq 4 \text{ and } x^2 > 2$ and $B = \{ x \in \mathbb{Q}: 0 \leq x \leq 4 \text{ and } x^2 < 2\}$.  Then both $A$ and $B$ are closed, because their complements in $\mathbb{Q}$ are open. However, neither is sequentially compact.  (This example also shows that adding a hypothesis that $B$ is bounded cannot substitute for sequential compactness.)
