You can do:

  if ('error' in result) {
    console.error(result.error); // ok: result inferred as 'Failure' 
  } else {
    doThings(result.value) // ok: result inferred as 'Success<T>'
  }

Even though you used string there, it's pretty type-safe because if you have a typo in your string, the inferred types will propagate other type errors

  if ('errors' in result) { // typo
    console.error(result.errors); // Type error: result inferred as 'never' 
  } else {
    doThings(result.value) // Type error: result inferred as Success<T> | Failure
  }

The discriminated union pattern is better [1]

    type Success<T> = { type: 'success', value: T};
    type Error = { type: 'error', value: string };
    type Result<T> = Success<T> | Error;

    if (result.type === 'error') {
      result.error;
    } else {
      result.value;
    }

[1] https://www.typescriptlang.org/docs/handbook/advanced-types....

...

What do you expect here? The compiler is doing exactly what is expected.