It's evidently a subspace since if
then
so that
. To prove it's closed first recall that the evaluation linear functional
is continuous since
Then, note that since
is a metric space singletons are closed and since the preimage of closed sets under continuous maps is closed it follows that
is closed. Note though that
.
For more general cases it can be more difficult. Showing that it's a subspace is usually simple, and showing it's closed is just usual topology/analysis.
Ask if you have more questions.