Just to clarify on relation to the title, we aren't using the t-distribution to estimate the mean (in the sense of a point estimate at least), but to construct an interval for it.
But why use an estimate when you can get your confidence interval exactly?
It's a good question (as long as we don't get too insistent on 'exactly', since the assumptions for it to be exactly t-distributed won't actually hold).
"You must use the t-distribution table when working problems when the population standard deviation (σ) is not known and the sample size is small (n<30)"
Why don't people use the T-distribution all the time when the population standard deviation is not known (even when n>30)?
I regard the advice as - at best - potentially misleading. In some situations, the t-distribution should still be used when degrees of freedom are a good deal larger than that.
Where the normal is a reasonable approximation depends on a variety of things (and so depends on the situation). However, since (with computers) it's not at all difficult to just use the $t$, even if the d.f. are very large, you'd have to wonder why the need to worry about doing something different at n=30.
If the sample sizes are really large, it won't make a noticeable difference to a confidence interval, but I don't think n=30 is always sufficiently close to 'really large'.
There is one circumstance in which it might make sense to use the normal rather than the $t$ - that's when your data clearly don't satisfy the conditions to get a t-distribution, but you can still argue for approximate normality of the mean (if $n$ is quite large). However, in those circumstances, often the t is a good approximation in practice, and may be somewhat 'safer'. [In a situation like that, I might be inclined to investigate via simulation.]