If you roll two fair six-sided dice, what is the probability that the sum is 4 or higher?

It is wrong because it is not $11$ equally possible outcome.

There is exactly $1$ way to get the sum to be $2$. ($1+1=2$)

but there is more than one way to get $3$. ($1+2=3, 2+1=3$)

As pointed out by others, the possible sums of the dice don't all have an equal probability of showing up. To see this, you can write it all out:

Die 1 | Die 2 | result
  1       1       2
  1       2       3
  1       3       4
  ...
  6       5      11
  6       6      12

When you look at the resulting table, there are 36 combinations. 1 of those is a 2 (ie you have a 1 in 36 chance of getting a 2 from D1 + D2), 2 of those are '3' etc.

Now it's easy to see how to get the chance of getting a 4 or higher.

$\frac9{11}$ is wrong precisely because it assumes the probabilities of getting each sum are equal. Here they are not: there's only one way to roll a sum of two (the snake eyes of gambling jargon) but six ways to roll a seven.