How do we know we cannot escape the event horizon?
Because a black hole is black. It's black because light can't get out. Light can't get out because at the event horizon the "coordinate" speed of light is zero. And we can't move faster than light. So we can't get out either.
A doubt assails me: deepening the Schwarzschild and Kerr space-times I found a beautiful article of the University of Rome, which provides a method to identify an event horizon based on the analysis of the square norm of constant "radial" coordinate hypersurfaces normal vector ¹ ².
It might be beautiful, but it's wrong. It says this:
"The metric, in these coordinates, is singular at r = 0 and at r = 2M. However, the singularity r = 2M is only an artifact of the coordinate choice, and can be removed by changing coordinates (and then it is
called “coordinate singularity”); the singularity r = 0, instead, is a true singularity of the metric (and is called “curvature singularity”)".
Why is it wrong? See Einstein's 1939 paper on a stationary system with spherical symmetry consisting of many gravitating masses. He said “g44 = (1 – μ/2r / 1 + μ/2r)² vanishes for r = μ/2. This means that a clock kept at this place would go at the rate zero". He also said “In this sense the sphere r = μ/2 constitutes a place where the field is singular”. Do you know how the singularity at the event horizon is removed? By using seconds of infinite length! I'm afraid it's "unphysical". See the Wikipedia article on Eddington-Finkelstein coordinates. They were invented by Roger Penrose. Then take a look at the section on tortoise coordinates.
In particular, taking advantage of the Strong Equivalence Principle, it is easily demonstrated that if the normal vectors are timelike the tangent vectors are spacelike i.e. once we enter the horizon we can no longer go out.
This is unphysical too. See what Einstein said in 1920: “the curvature of light rays occurs only in spaces where the speed of light is spatially variable”. At the event horizon, the speed of light is zero, and it can't go lower than that. Hence it's clear that Oppenheimer and Snyder’s 1939 “frozen star” paper on continued gravitational contraction describes the black hole better than your article. It's like Ruffini and Wheeler said in their 1971 article introducing the black hole, “in this sense the system is a frozen star”. That means the r=0 curvature singularity doesn't exist.
As for the strong equivalence principle, see the mathspages article on the many principles of equivalence where you can read this: “the modern statement of the strong equivalence principle, of the assertion that the laws of physics are the same for all frames of reference (i.e. independent of velocity) is also conceptually quite distinct from the original meaning of Einstein’s equivalence principle”. The strong equivalence principle is nothing to do with Einstein, and gamma ray bursters make it clear that it doesn't apply absolutely. So you shouldn't use it to speculate about the interior of a black hole.
The question is: since gravity is a geometric property of space-time, how can we be sure that by heavily disturbing the region of the event horizon this effect is not destroyed i.e. it is possible to escape from the horizon itself?
Because gravity isn't a geometric property of space-time. You won't find Einstein saying that. Instead you'll find him saying a gravitational field is a place where space is neither homogeneous nor isotropic. That's caused by a concentration of matter in the guise of a massive star "conditioning" the surrounding space. Because of this the speed of light varies. Look again at this: “the curvature of light rays occurs only in spaces where the speed of light is spatially variable”. Light doesn't curve because it follows a geodesic. It doesn't follow the curvature of spacetime. It curves because the speed of light varies. The important thing to note about all this, is this: at the event horizon the speed of light is zero, and it can't go lower than that. So there's no gradient in the speed of light, and no gravitational field.
Let me explain better: by introducing any energy source, gravity changes i.e. space-time changes. Considering that the horizon manifests itself through a geometric property (gravity), how can we be sure that the prediction "is impossible to get out of an event horizon" is valid even if on the horizon there is a star, an asteroid or, in general, an object with non-negligible energy effects capable of significantly altering the geometry of space-time?
Because when you add a concentration of energy to a region of space, you alter the surrounding space such that you set up a gradient in the speed of light. That's what a gravitational field is - a region of space where the speed of light is spatially variable. Light goes slower at a lower elevation. That's why optical clocks run slower when they're lower. However clocks can't go slower than stopped. The event horizon is a place where the speed of light is zero, and you can't make it go slower than that.
On page 50 of your second reference the author says "this situation is generally considered unphysical". When you read the original general relativity in the Einstein digital papers, you come to realise that some other situations invented in the 1960s are unphysical too.