Does there exist a space which has trivial fundamental group but is not path connected? I am asking this because the definition of space X to be simply connected is that X is path connected and has trivial fundamental group. I am not able to understand the importance of space X to be path connected in this definition.
 A: The assumption of path-connectivity in the definition of simple-connectedness is one of convenience, for two reasons.


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*As mentioned, this makes the identification of the fundamental group independent of choice of basepoint (necessary for its construction), up to isomorphism.  Since working up to isomorphism is "good enough" for 99% of algebraic arguments, this is a convenient thing to do.

*Once you get further along you will see how to use constructs called covering spaces and fiber bundles which are quite powerful.  Some of the technical theorems require certain niceness of the spaces or basepoints involved, and assuming path-connectedness either makes things nice or makes the ugliness manageable so that you still get your theorem.
And, to answer the question in your title, take any two simply-connected spaces $X$ and $Y$ and form their disjoint union $X \amalg Y$.  The result is not path-connected, but for any choice of basepoint $z_0$ we have $\pi_1(X \amalg Y, z_0)$ trivial.  For a concrete instance, consider the union of two (distinct) parallel lines in the Euclidean plane, with the ordinary subspace topology
A: Yes such spaces exist. One example which can be found in Munkres Topology 2nd Edition is the dictionary order topology (Example 2, p. 85) on the unit square (Example 3, p. 90). This is an example of a space that is connected: if it were the disjoint union of two open sets, then the interval [a,b] would be too. But this space is not path connected (the only path connected components of it are the vertical intervals).
It thus has trivial fundamental group, since $\pi_1(X,x_0) \simeq \pi_1(I)$ where $I$ denotes the interval $[0,1]$.
Topology is full of counterexamples. There's an entire book of them: Counterexamples in Topology Steen & Seebach (1995).
In the case of simple connectedness, part of the reason for the path-connected assumption is to rule out pathological spaces such as the one above.
