Type inference only works on the same line as the variable declaration

I ran into this odd behavior while writing some tests, that do the typical describe, declare a closure-scope variable, and in beforeEach, assign some object to that variable that is used in each of the tests inside.

Search Terms:
infer, inference, closure, scope, broken, any, cast

Initially, I thought it was limited to closure scope, but I then started reducing the reproduction sample more and more, and realized, it's broken even if it's on the very next line:

Code

// given
interface Foo {
    id: string;
}

function make(): Foo {
    return {
        id: 'akatechis',
    }
}

// u1 is correctly inferred to be of type Foo.
let u1 = make();

// u2 is incorrectly inferred to be of type any.
let u2;
u2 = make();

// this was the scenario in which I originally ran into this:
// a closure scoped variable that is assigned in an inner function
function outer () {
    // u3 is inferred to be of type any
    let u3;
    return {
        inner() {
            // u3 here is still any
            u3 = make();

            // surprisingly, returning u3 here causes the return type of inner() to be Foo, not any???
            return u3;
        }
    }
}

Expected type declarations:

interface Foo {
    id: string;
}
declare function make(): Foo;
declare let u1: Foo;
declare let u2: Foo; // <---- this should have been inferred to be Foo
declare function outer(): {
    inner(): Foo;
};

Actual type declarations:

interface Foo {
    id: string;
}
declare function make(): Foo;
declare let u1: Foo;
declare let u2: any;
declare function outer(): {
    inner(): Foo;
};

Playground Link: playground

Related Issues:
The only thing that I found somewhat close to this issue was the question in the FAQ about unused generic types causing type widening, but I don't think this is it, because I'm not using generics here.

Comienza con la reproducción vinculada en TypeScript Playground y compara las declaraciones inferidas para u1, u2 y outer(). Lee la entrada relacionada de las FAQ sobre type widening para descartar esa explicación y, después, identifica el comportamiento de inferencia de tipos responsable de la discrepancia. Se considera terminado cuando u2 se infiere como Foo, conservando el resultado mostrado de outer().