Let's go through this sentence-by-sentence.
You can't find a solid chunk (range of numbers) that does not contain a rational number.
This is true: The rational numbers are dense in the reals, so every segment $[a, b]$ in the real numbers contains some rational number.
Which means that irrational numbers are just points on the number line, not lines.
This is half-true: Because of the first statement, there is no segment $[a, b]$ which is a subset of the irrationals, since all such segments contain at least one rational number. However, on the other hand, these irrational numbers are not just isolated points, meaning that in any segment $[a, b]$ which contains an irrational number $q$, there will always be another irrational number $p\neq q$ such that $p$ is also in the segment $[a, b]$.
Thus, it is wrong to think about the irrational numbers as just isolated points on the number line, because they aren't: Even though there is no segment consisting of entirely irrational numbers, there are still irrational numbers which are arbitrarily close to each other.
Now, you could also say the same thing about the rationals: Even though there is no segment consisting of entirely rational numbers, there are still rational numbers which are arbitrarily close to each other. Thus, whether or not a set can be thought of as a bunch of isolated points or not doesn't really have anything to do with if the set is countable or not. This gets us to the third statement:
Which means you just need to name all the points to count the irrationals.
This statement is false, and does not follow from the above two statements because of the argument I have presented above. Just because there is no continuous segment which is a subset of the irrationals does not automatically mean you can make a list all of the irrational numbers by just listing all of the points. This sentence here is pretty vague and does not highlight any way to actually create a bijection between the set of natural numbers and the set of irrational numbers, which is what it really means to be a "countable set." Thus, this proof is invalid because of this vague third statement, which does not actually construct a clear way to make a list of all the irrational numbers.