A constant rate of events per unit time is called a Poisson process when the outcomes in one time interval are independent of the outcomes in any other time interval. This independence assumption is usually valid and supported by physical considerations. When not, it can be tested.
A Poisson process is characterized by a single parameter. The rate per unit time is usually chosen. Let's call it $\lambda$. This means the expected number of events observed throughout a duration $dt$ is $\lambda dt$. The variance of the number of events is also equal to $\lambda dt$.
Suppose you were to split the duration $dt$ into nonoverlapping subintervals $dt_1, dt_2, \ldots, dt_k$, with $dt_1 + dt_2 + \cdots + dt_k = dt$. Then:
The expected number of observations would equal the sum of the expectations,
$$\sum_{i=1}^k \lambda dt_i = \lambda \sum_{i=1}^k dt_i = \lambda dt,$$
just as it should.
The variance of the number of observations would equal the sum of the variances because the numbers of observations are independent. The calculation is exactly the same, only this time the quantities represent the variances:
$$\sum_{i=1}^k \lambda dt_i = \lambda \sum_{i=1}^k dt_i = \lambda dt.$$
Therefore, splitting the data into groups of observations does not improve the precision with which you can estimate the rate.
It can pay to go a little further. Suppose the rate does vary over time, but slowly enough that assuming a constant rate within each subinterval $dt_i$ is a good approximation. Let $X_i$ have Poisson$(\lambda_i)$ distributions and be independent, just as before. (To simplify the notation, I have incorporated the dependence on the durations $dt_i$ within the parameters $\lambda_i$.) Upon observing each of the $X_i$ you might want to estimate the underlying mean rate (per smaller interval) as the sample mean
$$\hat\lambda = \frac{1}{k} \sum_{i=1}^k x_i$$
and the variance in that rate as the sample variance
$$\hat\sigma^2 = \frac{1}{k-1} \sum_{i=1}^k (x_i - \hat\lambda)^2.$$
Using the Poisson distribution facts that $\mathbb{E}(X_i) = \lambda_i$ and $\mathbb{E}(X_i^2) = \text{Var}{X_i} + \mathbb{E}(X_i)^2 = \lambda_i + \lambda_i^2$, a little algebra shows that
$$\mathbb{E}(\hat\lambda) = \frac{\lambda}{k}$$
($\lambda = \sum_i \lambda_i$), which we may interpret as the mean rate per (smaller) interval, and
$$\mathbb{E}(\sigma^2) = \frac{1}{k-1}\left(\sum_i (\lambda_i - \lambda/k)^2\right) + \frac{\lambda}{k}.$$
The right hand term is the variance of a Poisson process of constant rate $\lambda/k$. The other term is the variance of the separate rates. When there is no variation, the first term drops out and this reduces to the previous result: the variance equals the mean and from that we derive the standard error as usual. But when the underlying rate does vary, its variance contributes to the expected sample variance, creating an overdispersed dataset.
Incidentally, the effect of the subdivision is never to decrease the estimated variance (and therefore the standard error): it can only increase it.
We see from this that the potential advantage of dividing the whole time interval into little parts is that it allows us to detect variations in the underlying rate and to use them to increase our estimate of the standard error. If you are confident that the underlying rate does not appreciably change, then there is no need to collect data from subintervals--but it couldn't hurt.
