$\lim \limits_{a \to 0} \frac{1}{a}[u(\frac{t}{a}+\frac{1}{2})-u(\frac{t}{a}-\frac{1}{2})]= \delta(t) $
The graph of $F_a(t)$ is a rectangle with base length $a$ and height $a^{-1}$. The area of the rectangle is therefore equal to $a \cdot a^{-1}=1$. The Dirac delta distribution is often defined as a certain (see below) limit of a sequence of suitable functions of integral 1 in the limit of vanishing support. This definition is already very intuitive, so that your intuition criterium should be met.

A more rigorous look uses a different definition of the delta distribution through the equation

$$\int_{-\epsilon}^{+\epsilon} f(x) \delta(x) dx = f(0)$$

for any open neighbourhood of radius $\epsilon$ around $0$ on which $f$ is smooth, as well as the condition $\delta(x\neq 0)=0$.

If you plug in your $F_a(x)$ in place of $\delta(x)$ we get

$$\int_{-\epsilon}^{+\epsilon} f(x) F_a(x) dx$$

Now the limit of $F_a$ for $a\to0$ is not to be understood of a limit in the usual function sequence sense. That limit does not exist in terms of a function. Rather, the limit is meant to be taken outside an integral under which $F_a$ appears. So taking the limit happens like

$$\lim_{a\to0}\int_{-\epsilon}^{+\epsilon} f(x) F_a(x) dx$$

In order to understand the limit, we can start at any finite but small enough $a$. If we choose $a&lt;2\epsilon$ and use that $F_a$ and with it the function under the integral vanishes outside of $[-a/2,a/2]$ and offer no contribution to the integral, the boundaries can be adjusted accordingly.

$$\lim_{a\to0}\int_{-a/2}^{+a/2} f(x) F_a(x) dx$$

Within these boundaries $F_a$ is constant and equals $1/a$. So the integral simplifies to

$$\lim_{a\to0} \frac{1}{a}\int_{-a/2}^{+a/2} f(x) dx$$

The requirements for $f$ are that it be smooth on the open interval of our integral. So we can expand $f(x)=f(0)+x \cdot f'(0) + \mathcal{O}(x^2)$ which together with the linearity of the integral results in

$$\lim_{a\to0} \frac{1}{a}\left( f(0) \int_{-a/2}^{+a/2} dx + f'(0) \int_{-a/2}^{+a/2} x dx + \int_{-a/2}^{+a/2} \mathcal{O}(x^2) dx\right)$$

The first integral contributes $a$, the second integral vanishes because of symmetry of the boundaries and the odd function under the integral, and the third integral contributes $\mathcal{O}(a^3)$.

This gives 

$$\lim_{a\to0} \frac{1}{a}\left( f(0) a + \mathcal{O}(a^3)\right) = \lim_{a\to0} \left( f(0)  + \mathcal{O}(a^2)\right) = f(0) $$

This result tells us that your $F_a$ works as a $\delta$ distribution if the limit is taken outside the integral. Or in other words, the "limit" of the function sequence is a delta distribution.

Edit: I just noticed I did not go into the second condition of $\delta(x \neq 0) = 0$. To show this for $F_a(x)$, just observe that for any $\hat{x}\neq0$ you can find an $\hat{a}$ so that $F_a(\hat{x})=0$ for all $a&lt;\hat{a}$.

Edit 2: Why is the condition $\delta(x \neq 0) = 0$ necessary at all? This one is a bit subtle. The integral condition should uniquely define the distribution since it must hold for all test functions $f$ from the space of locally smooth functions. The problem is that the test functions are not enough to rule out distributions that are identical to $\delta$ only almost everywhere. For example, if we had a distribution $\bar{\delta}$ that coincides with $\delta$ with exception of $\bar{\delta}(1)=1$, then the integral condition would still be met because the set that supports the deviation is of measure zero. The requirement of vanishing $\delta$ for non-zero arguments excludes these cases. It is however often omitted because it does not present a problem unless you let the delta distribution act on other distributions.