dict((k, v) for k, v in points.items() if all(x < 5 for x in v))

You could choose to call .iteritems() instead of .items() if you're in Python 2 and points may have a lot of entries.

all(x < 5 for x in v) may be overkill if you know for sure each point will always be 2D only (in that case you might express the same constraint with an and) but it will work fine;-).

Why are you iterating over your dict? That defeats the purpose. Just do

[d for d in l if key in d]

Or if you feel some compulsion to use lambda and filter:

filter(lambda d: key in d, l)

Note, these return lists instead of the corresponding value. If you want the value, you'll have to index into the list. Simply out, using a loop is probably the most reasonable approach, just don't iterate over your dict:

def f(key, l):
    for d in l:
        if key in d:
            return d[key]

As an alternative leveraging your existing list statement, you could add a map statement that grabs the employedId key:

from operator import itemgetter

list(map(itemgetter('employedId'), filter(lambda tag: tag['age'] == 22, ages)))
[1, 6]

Otherwise, I think @CoryKramer's answer is a bit more readable

try

y = filter(lambda x:dict[x] > 0.0,dict.keys())

the lambda is feed the keys from the dict, and compares the values in the dict for each key, against the criteria, returning back the acceptable keys.