Others have dealt with the specific error in your reasoning. I want to point out that the problem is not well-defined: without more information it is impossible to say what the probability is that both children are girls. In particular, it not true that the answer is definitely $1/3$.
Consider the following three scenarios. In each of them I choose one family at random from the pool of all families with exactly two children, in such a way that each such family is equally likely to be chosen, and then I make one of the following two statements to you:
Statement 1: At least one of the children is a girl.
Statement 2: At least one of the children is a boy.
Scenario A: If the older child is a girl, I make Statement 1; if the older child is a boy, I make statement 2.
Scenario B: If both children are girls, I make Statement 1; otherwise, I make Statement 2.
Scenario C: If both children are boys, I make Statement 2; otherwise, I make Statement 1.
Suppose that I’ve made Statement 1: I’ve told you that at least one of the children is a girl. What is the probability that both are girls?
Scenario A: The older child is definitely a girl, so the probability that both are girls is simply the probability that the younger child is a girl, which is $1/2$.
Scenario B: Both children are definitely girls, or I’d have made Statement 2, so the probability that both are girls is $1$.
Scenario C: The only possibility that is ruled out is that both are boys. Three possibilities remain: both are girls; the elder is a girl and the younger a boy; the elder is a boy and the younger a girl. These are equally likely, so the probability that both are girls is $1/3$.
In short, the probability depends on what I would have said had I chosen a different family. Using more complicated decision rules, I could arrange scenarios in which the probability that both children are girls has other values. Here’s an example:
Scenario D: I roll a fair die. If both children are girls, I make Statement 1. If both are boys, I make Statement 2. If one is a boy and the other a girl, I make Statement 1 if the die comes up $6$; otherwise, I make Statement 2.
There are $6\cdot 2\cdot 2=24$ equally likely events, one for each combination of die roll, sex of older child, and sex of younger child. I’ll represent a specific combination by a string like $3BG$, meaning that I rolled a $3$, the elder child is a boy, and the younger child is a girl. The fact that I made Statement 1 rules out six of these $24$ events: $1BB,2BB,3BB,4BB,5BB$, and $6BB$. It also rules out all five $nBG$ events in which $n\ne 6$ and all five $nGB$ events in which $n\ne 6$. The only possible events, therefore, are the six $nGG$ events, $6BG$, and $6GB$. In six of these eight equally probable cases both children are girls, so the probability that both are girls in this scenario is $6/8=3/4$.