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.

リンクされている TypeScript Playground の再現コードから始め、u1、u2、outer() について推論された宣言を比較します。type widening に関する関連 FAQ エントリを読んでその説明を除外し、次に不一致の原因となっている型推論の挙動を特定します。u2 が Foo として推論され、示されている outer() の結果が維持されれば完了です。