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The fact that any composition of isometries is also an isometry is true and quite obvious.
The proof is a trivial application of the fact that equality and congruence are transitive.
Suppose that we have a segment which is mapped by an isometry to , which is mapped by another isometry to .
Thus, we have the following: Since is an isometry, .
Since is an isometry, .
By the Transitive Property of Equality, .
Since was an arbitrary segment, we can see that the following argument holds for all segments.
Thus, the composition preserves distances, so it is an isometry.