Proof of Dirac Delta's sifting property

Well, as you mention, no truely rigorous treatment can be given with such a description of the Delta Dirac function - no such function actually satisfies those requirements. Thus, I won't take too much effort to make the below too precise.

Where it says "sufficiently smooth", it doesn't actually need anything there at all! Whatever $f$ is, as long as it is finite almost everywhere, the product with that delta function will be $0$ away from a neighbourhood of $t$, so you can restrict the integral like that.

For the extraction of $f$, being continuous in a closed neighbourood of $x=t$ is enough. If $f$ is continuous through $ [ t-\epsilon, t+ \epsilon ] $ then by the Extreme value theorem it attains a maximum and minimum in that interval, call them $ M $ and $m$ respectively.

Then since $ m \leq f(x) \leq M$ is that range, $$ m=\int^{t+\epsilon}_{t-\epsilon} m \delta(x-t) dx \leq \int^{t+\epsilon}_{t-\epsilon} f(x) \delta(x-t) dx \leq \int^{t+\epsilon}_{t-\epsilon} M \delta(x-t) dx = M.$$

Now as $\epsilon \to 0$, both $m$ and $M$ go to $f(t)$ as $f$ is continuous so by the Squeeze theorem, $$ \int^{t+\epsilon}_{t-\epsilon} f(x) \delta(x-t) dx \to f(t) $$ as $\epsilon \to 0$.

Theory of Distributions by J. Ian Richards and Heekyung Youn gives rigorous proofs of things like this without need for functional analysis, topology, or measure theory.

Definitely the proof you give is extremely "hand-wavy". In particular, the two defining properties you give cannot be taken literally if one conceives of a function as something where you put in a number $x$ and get out a number $f(x)$. To take those two properties is intuitive but very very "hand-wavy".