Dirac delta function as a limit of sinc function

The Dirac delta is to be defined as a distribution: a linear functional acting on the space of smooth compactly supported functions. So this limit is to be understood as:

$$\lim_{\varepsilon \to 0^+} \int_{-\infty}^\infty \frac{\sin \left ( \frac{x}{\varepsilon} \right )}{\pi x} f(x) dx = f(0)$$

whenever $f$ is smooth and has compact support. (Actually, the Dirac delta may be extended to continuous compactly supported functions, but this is the starting point.)

That said, the argument here is similar to the usual situation of showing that a certain family of functions is an approximate identity. First you should rewrite in a more convenient form:

$$g_\varepsilon(x) \equiv \frac{\sin \left ( \frac{x}{\varepsilon} \right )}{\pi x} = \frac{1}{\varepsilon} \frac{\sin \left ( \frac{x}{\varepsilon} \right )}{\pi \frac{x}{\varepsilon}}.$$

Since $\int_{-\infty}^\infty \frac{\sin(x)}{x} dx = \pi$, we see that $g_\varepsilon$ has integral $1$. The next condition to be checked is that for any compact set $K$ and any $\delta > 0$:

$$\lim_{\varepsilon \to 0^+} \int_{K \setminus (-\delta,\delta)} g_\varepsilon(x) dx = 0.$$

If you haven't seen this before, the intuition here is that $g_\varepsilon$ has the same mass for any $\varepsilon$ but that this mass is becoming concentrated in an arbitrarily small interval around $0$.

Most commonly, we choose $g_\varepsilon$ to be absolutely integrable on the whole real line, and in this case we could look at $\mathbb{R} \setminus (-\delta,\delta)$ instead. In this situation, we don't have absolute integrability. But if we only work with test functions with compact support then we can make this generalization to get things to work out, by taking $K$ to be the support of $f$ in the hypothesis above.

A related proof is by Fourier transforms. Here's a sketch of this proof:

The sinc function (with appropriate scaling) is the Fourier transform of the indicator function of an interval centered at $0$. The delta function is the Fourier transform of the constant function $1$ (again with appropriate scaling). When $\varepsilon$ is small (when the mass has gotten concentrated), $f_\varepsilon$ is the Fourier transform of an indicator function of a large interval centered at zero. (The small/large phenomenon here is more generally the uncertainty principle in Fourier analysis.) Since the indicator functions are converging to the constant function $1$, their Fourier transforms are converging to the Fourier transform of $1$, i.e. $\delta$.

This proof with Fourier transforms is harder to formalize. For instance, we need to show that the Fourier transform is continuous in some appropriate sense. But it may be more intuitive.

In addition to the answer by Ian, one very important lemma you need to prove the second part, i.e $$\lim_{\varepsilon \to 0^+} \int_{K \setminus (-\delta,\delta)} g_\varepsilon(x) dx = 0.$$ is the Riemann-Lesbegue lemma which shows how an increasingly oscillatory function under integration approaches zero measure. You can find it mentioned here :- https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470723876.app3(Page 498)