I would say that no, it is not true.

An idea for the example is as follows. Imagine that you are having 100 intervals of length 1/100, 100000 of length 1/10000, 10^6 of length 1/10^6, etc. The sum of their lengths clearly diverges. Now, what happens for the set left?
I am going to modify your process: imagine that the intervals of length 1/N have only discrete allowed positions, $[\frac{k}{N},\frac{k+1}{N}]$, where k=0,1,...,N-1.

It is, formally speaking, a different process and a different question. However, they seem sufficiently alike (and I would say that there is a way to deduce the conclusion for the original one).

Now, for this modified process, if you are considering a given interval of length 1/N at the time when N intervals of length 1/N are being placed, it has a probability of survival $(1-1/N)^N \approx 1/e$. And though it is not independent for different intervals, I guess we can safely say that about $1/e$-th proportion of such intervals survives.

The intervals come in waves: the first wave of N=100 intervals of length 1/100, the second of $N=10^4$ intervals of length $1/10^4$, etc. Each interval that has survived one wave is decomposed into a 100 intervals that face the next one. And about $1/e$-th proportion of them survive.

So there is an (almost: remember, no independence!) Galton-Watson tree with the expectation of $100/e>1$ descendants of each interval, and hence with positive probability there is an exponential growth of number of surviving intervals, in the limit defining a Cantor set that in a complement to the union of intervals placed.

Finally, to get back to the original process, you can consider that an interval $[\frac{k}{N},\frac{k+1}{N}]$ is removed from the complement once it is *intersected* by one of the randomly placed intervals of length $1/N$. Then each interval still has an $\approx 1/e^2$ chance of survival, and I'm sure whichever way you use to handle the above argument formally, it will still be applicable here.