Probability of selecting an even number from the set of natural numbers In the above problem, if I form the sample space as the following set:
$$S= \{\text{even number, odd number}\}.$$
Obviously both events are equally likely, since there is no reason to prefer one over the other. Also, this is the set of all possible outcomes of the experiment in which a number is being randomly selected from the set of natural numbers (note that an experiment can have more than one sample spaces). So why is $P(\text{selecting an even number})$ not as simple as
$$\frac{1}{2}\cdot \frac{\text{number of favourable outcomes from the sample space}}{\text{total outcomes in sample space}}?$$
 A: There is no uniform distribution on a countably infinite set, so you can't define a probability like this just by counting.
There are a couple of ways people get around this.  The most natural is to introduce a cutoff $N$ and then pass to the limit as $N\to \infty$.  Thus, for each natural number $N$, we can consider the integers between $-N$ and $N$.  Of course we can count those and we can define $P_N$ to be the probability that a (uniformly) randomly selected integer in this interval is even.  We can then define $P=\lim_{N\to \infty}P_N$.  It's not difficult to see that this implies $P=\frac 12$, as you would expect.  See Natural Density.
Another method is to consider the sums of the reciprocals of the integers you are trying to "count" (taking care to remove $0$ from the list).  This is more technical and a lot less intuitive but it has some analytic properties that sometimes make it easier to work with.  This method comes up when considering subsets of the primes (see Analytic Density).  It gives $\frac 12$ in this case as well.  As you suggest, any "natural" method for computing this probability really ought to yield $\frac 12$ by symmetry.
