This comes from the fact that $\newcommand{\Var}{\operatorname{Var}}\newcommand{\Cov}{\operatorname{Cov}}\Var(X+Y) = \Var(X) + \Var(Y) + 2\cdot\Cov(X,Y)$ and for a constant $a$, $\Var( a X ) = a^2 \Var(X)$.
Since we are assuming that the individual observations are independent the $\Cov(X,Y)$ term is $0$ and since we assume that the observations are identically distributed all the variances are $\sigma^2$. So
$\Var( \frac{1}{n} \sum X_i ) = \frac{1}{n^2} \sum \Var(X_i) = \frac{1}{n^2} \times \sum_{i=1}^n \sigma^2= \frac{n}{n^2} \sigma^2 = \frac{\sigma^2}{n}$
And when we take the square root of that (because it is harder to think on the variance scale) we get $\dfrac{\sigma}{\sqrt{n}}$.
More intuitively, think of 2 statistics classes: in the first the teacher assigns each of the students to draw a sample of size 10 from a set of tiles with numbers on them (the teacher knows the true mean of this population, but the students don't) and compute the mean of their sample. The second teacher assigns each of his/her students to take samples of size 100 from the same set of tiles and compute the mean. Would you expect every sample mean to exactly match the population mean? or to vary about it? Would you expect the spread of the sample means to be the same in both classes? or would the 2nd class tend to be closer to the population? That's why it makes sense to divide by a function of the sample size. The square root means we have a law of diminishing returns, to halve the standard error you need to quadruple the sample size.
As for the name, the full name is "The estimated standard deviation of the sampling distribution of x-bar"; it only takes saying that a few times before you appreciate having a shortened form. I don't know who first substituted "error" for "deviation" this way, but it stuck. The standard deviation measures variability of individual observations; the standard error measures variability in estimates of parameters (based on observations).
There are at least three basic problems which can readily be explained to beginners:
The "new" SD is not even defined for infinite populations. (One could declare it always to equal zero in such cases, but that would not make it any more useful.)
The new SD does not behave the way an average should do under random sampling.
Although the new SD can be used with all mathematical rigor to assess deviations from a mean (in samples and finite populations), its interpretation is unnecessarily complicated.
1. The applicability of the new SD is limited
Point (1) could be brought home, even to those not versed in integration, by pointing out that because the variance clearly is an arithmetic mean (of squared deviations), it has a useful extension to models of "infinite" populations for which the intuition of the existence of an arithmetic mean still holds. Therefore its square root--the usual SD--is perfectly well defined in such cases, too, and just as useful in its role as a (nonlinear reexpression of) a variance. However, the new SD divides that average by the arbitrarily large $\sqrt{N}$, rendering problematic its generalization beyond finite populations and finite samples: what should $1/\sqrt{N}$ be taken to equal in such cases?
2. The new SD is not an average
Any statistic worthy of the name "average" should have the property that it converges to the population value as the size of a random sample from the population increases. Any fixed multiple of the SD would have this property, because the multiplier would apply both to computing the sample SD and the population SD. (Although not directly contradicting the argument offered by Alecos Papadopoulos, this observation suggests that argument is only tangential to the real issues.) However, the "new" SD, being equal to $1/\sqrt{N}$ times the usual one, obviously converges to $0$ in all circumstances as the sample size $N$ grows large. Therefore, although for any fixed sample size $N$ the new SD (suitably interpreted) is a perfectly adequate measure of variation around the mean, it cannot justifiably be considered a universal measure applicable, with the same interpretation, for all sample sizes, nor can it correctly be called an "average" in any useful sense.
3. The new SD is complicated to interpret and use
Consider taking samples of (say) size $N=4$. The new SD in these cases is $1/\sqrt{N}=1/2$ times the usual SD. It therefore enjoys comparable interpretations, such as an analog of the 68-95-99 rule (about 68% of the data should lie within two new SDs of the mean, 95% of them within four new SDs of the mean, etc.; and versions of classical inequalities such as Chebychev's will hold (no more than $1/k^2$ of the data can lie more than $2k$ new SDs away from their mean); and the Central Limit Theorem can be analogously restated in terms of the new SD (one divides by $\sqrt{N}$ times the new SD in order to standardize the variable). Thus, in this specific and clearly constrained sense, there is nothing wrong with the student's proposal. The difficulty, though, is that these statements all contain--quite explicitly--factors of $\sqrt{N}=2$. Although there is no inherent mathematical problem with this, it certainly complicates the statements and interpretation of the most fundamental laws of statistics.
It is of note that Gauss and others originally parameterized the Gaussian distribution by $\sqrt{2}\sigma$, effectively using $\sqrt{2}$ times the SD to quantify the spread of a Normal random variable. This historical use demonstrates the propriety and effectiveness of using other fixed multiples of the SD in its stead.
Best Answer
The standard deviation calculated with a divisor of $n-1$ is a standard deviation calculated from the sample as an estimate of the standard deviation of the population from which the sample was drawn. Because the observed values fall, on average, closer to the sample mean than to the population mean, the standard deviation which is calculated using deviations from the sample mean underestimates the desired standard deviation of the population. Using $n-1$ instead of $n$ as the divisor corrects for that by making the result a little bit bigger.
Note that the correction has a larger proportional effect when $n$ is small than when it is large, which is what we want because when n is larger the sample mean is likely to be a good estimator of the population mean.
When the sample is the whole population we use the standard deviation with $n$ as the divisor because the sample mean is population mean.
(I note parenthetically that nothing that starts with "second moment recentered around a known, definite mean" is going to fulfil the questioner's request for an intuitive explanation.)