Explanation of `let` and block scoping with for loops

Is this just syntactic sugar for ES6?

No, it's more than syntactic sugar. The gory details are buried in §13.6.3.9 CreatePerIterationEnvironment.

How is this working?

If you use that let keyword in the for statement, it will check what names it does bind and then

• create a new lexical environment with those names for a) the initialiser expression b) each iteration (previosly to evaluating the increment expression)
• copy the values from all variables with those names from one to the next environment

Your loop statement for (var i = 0; i < 10; i++) process.nextTick(_ => console.log(i)); desugars to a simple

// omitting braces when they don't introduce a block
var i;
i = 0;
if (i < 10)
    process.nextTick(_ => console.log(i))
    i++;
    if (i < 10)
        process.nextTick(_ => console.log(i))
        i++;
        …

while for (let i = 0; i < 10; i++) process.nextTick(_ => console.log(i)); does "desugar" to the much more complicated

// using braces to explicitly denote block scopes,
// using indentation for control flow
{ let i;
  i = 0;
  __status = {i};
}
{ let {i} = __status;
  if (i < 10)
      process.nextTick(_ => console.log(i))
      __status = {i};
}   { let {i} = __status;
      i++;
      if (i < 10)
          process.nextTick(_ => console.log(i))
          __status = {i};
    }   { let {i} = __status;
          i++;
          …

I found this explanation from Exploring ES6 book the best:

var-declaring a variable in the head of a for loop creates a single binding (storage space) for that variable:

const arr = [];
for (var i=0; i < 3; i++) {
    arr.push(() => i);
}
arr.map(x => x()); // [3,3,3]

Every i in the bodies of the three arrow functions refers to the same binding, which is why they all return the same value.

If you let-declare a variable, a new binding is created for each loop iteration:

const arr = [];
for (let i=0; i < 3; i++) {
    arr.push(() => i);
}

arr.map(x => x()); // [0,1,2]

This time, each i refers to the binding of one specific iteration and preserves the value that was current at that time. Therefore, each arrow function returns a different value.