Are $\{0\},\{1\}$ clopen in $\{0,1\}$ with the Euclidean metric? Are $\{0\},\{1\}$ clopen in $\{0,1\}$ with the Euclidean metric?
I think they are, but I would like a confirmation. Thank you.
 A: If you are using the Euclidean metric as you indicated in the comments then yes.
$N_{.25}(0) = \{x \in \{0,1\}| d(x,0) < .25\} = \{0\}$ so {0} is open.  Similar for {1}.  As closed means compliment is open, $\{0\}^c = \{1\}$ which is open so {0} is closed.  Similar for {1}.
But to be more general to declare a set open and closed one has to declare a topology on the space.  This is, more or less, arbitrarily defining certain sets to be open by fiat. {0,1} as a subset of real numbers (or any other space) with the euclidean (or any other) metric is simply one arbitrary topology.  We could just as easily declare a topology where neither {0} nor {1} are open or closed.  (Although in all topologies $\emptyset$ and the universal set are open.  For a set with two elements there are four possible topologies on whether we declare which of the proper subsets to be open or not.)
(So the four possible topologies are:A {0} and {1} are not open and not closed. This is called the trivial topology. B:  {0} is open but {1} isn't.  So {1} is closed but {0} isn't.  C: {1} is open but {0} isn't.  So {0} is closed by {1} isn't.  D: both {0} and {1} are open.  So both are closed.  This is equivalent to any metric one can place on {0,1}.)
(Postscript:  If {a,b} is any metric space, not just the euclidean, nor anesc. a subset of the reals,   then $c = d(a,b) > 0$ so for any $0 < \epsilon < c$ we have $B_{\epsilon}(x) = \{y \in\{a,b\}| d(x,y) < \epsilon\} = \{x\}$ so {x} is open.  So in any 2 element metric space all sets are clopen.  Actually in all finite metric spaces all sets are clopen.)
A: Yes. Both sets are clopen. To see this note that $\{0\}=B(0,\frac{1}{2})$ and $\{1\}=B(1,\frac{1}{2})$, both of which are open sets. So, $\{0\}$ and $\{1\}$ are both open. They are also both closed since they are complements of each other.
