Why, for Menger Sponge (3D generalization of Cantor Set and Sierpiński Triangle), there exists area surface? Shouldn't it be zero instead of infinite? Imagine that you have a cube with an edge length of $a$. Let's call this cube $S_0$.
Now, imagine that you virtually sub-divide it into $27$ smaller cubes, each with an edge's length of $a/3$. Remove the small cube from the middle of each of $S_0$'s face, as well as the small cube containing the center of $S_0$, removing a total of 7 smaller cubes.
Doing this, we got a cube with a "tunnels" carved inside it, let's call it $S_1$.

The surface area of our figure increased by $1/3$-th, that is from $6a^2$ to $8a^2 = 6a^2 \cdot \frac{8}{9} + \frac{24}{9}a^2$, as we cut out $1/9$-th of each face, but added $24$ small squares inside by carving out the tunnels.
Volume of $S_1$ is just $\frac{20}{27}a^3$.

Now, we can continue the process of carving out our cube and create second iteration of Menger Sponge $S_2$, for which we will essentially be replacing each of $20$ smaller cubes (actually not cubes anymore) with a smaller (scaled by $1/3$) copy of $S_1$.

In general, for any $n\in N$ greater than $0$, we can now create $S_{n+1}$ out of $S_n$ by replacing each of $20$ carved cubes that are consisting on $S_n$ with a scaled by $1/3$ copy of $S_n$. We can also calculate that:

*

*Volume of $S_n$: $\quad \quad \quad V_n = a^3 (\frac{20}{27})^n \longrightarrow 0$,


*Surface area of $S_n$: $\ \quad  A_n = a^2 \Bigl(4(\frac{8}{9})^n+2(\frac{20}{9})^n \Bigr)\longrightarrow \infty $
if we take limit for $n \to \infty$.

However, we are still talking about finite "$n$-th iteration of the Menger Sponge".
To reach what we call a Menger Sponge (also Menger Cube, Sierpiński Cube etc.) we would need to repeat the process infinitely many times and take the limit of this process.
Now, even though the limit of the surface area $A_n$ goes to infinity, my intuition screams that surface area of Menger Cube should be zero. I mean, it's not like with Koch curve, where we keep getting longer and longer border line, which leads to "infinitely long" set of points that can still be considered a curve. In contrast, with Menger Sponge the area (the way I look at it) should not exist or it should be of measure zero, as we are practically "carving out" both everything that contains volume, as well as surface area.
However! If I try to search Google for 'menger sponge surface area', I am being hit with not just one, but multiple independent results stating things like "the Menger sponge has infinite surface area but zero volume", for example this article from Scientific American:
A Few of My Favorite Spaces: The Menger Sponge

Is my intuition wrong, as it can often err with some border or extreme cases in mathematics?
Even if for finite $n$-th iteration we can find the area by filling its surface with smaller and smaller 2-dimensional open balls of non-zero radius, can we even put a single one open ball on top of the Menger Sponge that is the limit construct using $S_n$?
If its area is non-zero, but infinite, then how do we define this area? How do we measure it, if we can't fit an "almost flat" 2D open ball anywhere on top of it, no matter how small finite radius it has?
Maybe I am not too rigorous in my explanation, but I think the idea is easy to understand:
I claim that there is no part of Menger Sponge that can be considered a surface with positive area.
Am I the one that's wrong here? Or it is internet that's trolling me? If it has no surface area (or it's zero), how can I prove it in nice and rigorous way?
 A: You should start by reexamining your definitions: You start with a compact subset $K$ of ${\mathbb R}^3$. Then you take a sequence of 3-dimensional nested solids $S_n$ (with "nice" boundary) such that
$$
\bigcap_{n\in {\mathbb N}} S_n=K.
$$
Then you say "take a limit $A$ as $n\to\infty$ of $Area(\partial S_n)$." You then declare this limit to be the "surface area" of $K$. The next question you should have asked yourself: Is this really a definition? What if one takes a different sequence of solids $S_n'$ satisfying all of the above conditions; is it true that
$$
\lim_{n\to\infty} Area(\partial S_n')=A ?
$$
One can show that in general the answer is negative: Depending on the choice of solids one can get infinite limits even though the original limit is finite. So, maybe your definition is not that good after all? Mathematicians struggled with such questions for awhile before converging on some common notions. I will not reproduce any of these here, the most commonly used notions are due to Hausdorff. Given a compact (let's restrict to compacts for simplicity) $K\subset {\mathbb R}^n$, one defines the Hausdorff dimension $d$  of $K$, which is a certain real number in the interval $[0,n]$. One also defines the $\alpha$-Hausdorff measure $\mu_\alpha(K)$ for each $\alpha\in [0,n]$. By the definition, $\mu_\alpha(K)=0$ whenever $\alpha>d$ and $\mu_\alpha(K)=\infty$ whenever $\alpha <d$. If you want to compute the "true volume" of $K$, you compute $\mu_d(K)$.
In order to understand the statements such as

"the Menger sponge has infinite surface area but zero volume"

you have to realize that these refer to $\mu_2(K)$ and $\mu_3(K)$. Since the Hausdorff dimension $d$ of the Menger curve is about 2.7, what they are saying is absolutely correct and your intuition, wherever it is coming from is plain wrong.
Now, regarding

I claim that there is no part of Menger Sponge that can be considered a surface with positive area.

Well, what exactly do you mean by a surface? As a topologist, I would say that a surface is a connected 2-dimensional manifold. If this is what you mean, then yes, Menger curve contains no nonempty surfaces whatsoever! This is because Menger curve has topological dimension 1 (not to be confused with the Hausdorff dimension), which is why topologists do not use the sloppy terminology "sponge." Spaces of topological dimension 1 contain no nonempty surfaces (since the latter have topological dimension 2).
Now, you might ask: How come Menger curve has topological dimension 1 and Hausdorff dimension $\approx 2.7$? The answer is that it is actually quite typical for fractals (whatever one means by this word) to have Hausdorff dimension strictly larger than topological dimension.
To your last question:

If it has no surface area (or it's zero), how can I prove it in nice and rigorous way?

As I said, the 2-dimensional Hausdorff measure (the "surface area") of the standard Menger curve $M$ is $\infty$, so you cannot prove that it is zero. To verify that it is $\infty$ you can read for instance the MA thesis (at the Wake Forest University) of Melissa Glass "Dimensions of self-similar fractals".
She computes the Hausdorff dimension ($\approx 2.7$) of the standard Menger curve $M$ on page 62. From this, it follows that $\mu_2(M)=\infty$.
