How to make a key optional?

There may be a more elegant way, but you can do it by removing C from the type and then adding it back as an optional property:

type TypeKey = 'A' | 'B' | 'C'
// Remove C  vvvvv−−−−−−−−−−−−−−−−−−−−−−−−−−−−vvvvvv
type Types = Omit<{ [key in TypeKey] : string}, 'C'> & {C?: string}
// Add it back as an optional property −−−−−−−−−−−−−−^^^^^^^^^^^^^^

let myTypes : Types = {
    A : 'Apple',
    B : 'Banna'
}

Playground link

In a comment you've asked:

it works, but any better way? something like type TypeKey = 'A' | 'B' | 'C?'

Not as far as I know, but you can turn this on its head and define TypeKey in terms of Types rather than the other way around:

interface Types {
    A: string;
    B: string;
    C?: string;
}

type TypeKey = keyof Types;

let myTypes : Types = {
    A : 'Apple',
    B : 'Banna'
}

Playground link

You can write it in more generic fashion as below:

type TypeKey = 'A' | 'B' | 'C'
type Types = { [key in TypeKey]: string }

type Optional<T, K extends keyof T> = Partial<Pick<T, K>> & Omit<T, K>

const myTypes: Optional<Types, 'C'> = {
  A: 'Apple',
  B: 'Banna'
}