How to extract single Type from a union?

Yes. Since TypeScript 2.8 supports conditional types you can do:

interface IFoo { 
    name: string
}

type FooArray = IFoo[] | number[] | undefined

type FindType<TWhere> = TWhere extends (infer U)[] ? (U extends object ? U : never) : never

type FoundType = FindType<FooArray> // FoundType == IFoo

Note that the U extends object ? U : never is required so that number is not matched.

For Completeness Exclude along with a type query can also be used resulting in something pretty readable

interface IFoo { 
    name: string
}

type FooArray = IFoo[] | number[]  | undefined

type ArrayValues = Exclude<FooArray, undefined>[number] // IFoo | number

type IFooExtracted = Exclude<ArrayValues, number> // IFoo