The answer by probably_someone is technically incorrect. He is confusing the Schwarzschild radius with an event horizon. They are not the same thing.
the conditions that the author assumed in the beginning are not
compatible with the conditions that admit the definition of a
Schwarzschild radius. The Schwarzschild solution of the Einstein field
equations requires that all of the mass of the universe is
concentrated in a physical singularity at $r=0$, and the rest is
vacuum. ...
Just because he calls it a Schwarzschild radius doesn't mean that it
is one.
A gravitational body does not need to be a black hole to have a Schwarzschild radius. Every massive body has a Schwarzschild radius, defined as $2GM/c^2$. For the Earth, the Schwarzschild radius is $2G(\text{Mass of Earth})/c^2$. This is the radius of the hypothetical sphere that you would have to compress the body's mass into for it to become a black hole. For the Earth, this is $9 \,\text{mm}$.
The Interior Schwarzschild metric is an example of a metric that uses the Schwarzschild radius and yet, this metric does not require a singularity at the center and is not a vacuum solution:
$$c^2 \mathrm{d}\tau^2 = \left(1 - \frac{r_s}{r}\right) c^2 \ \mathrm{d}t^2 - \left(1 - \frac{r_s}{r}\right)^{-1} \mathrm{d}r^2 - r^2(\mathrm{d}\theta^2 + \sin^2\theta \ \mathrm{d}\phi^2),$$
where $r$ is the surface of the fluid body. For a stationary particle at the surface of the fluid body, we get: $$\frac{\mathrm{d}\tau^2}{\mathrm{d}t^2} = \sqrt{\left(1 - \frac{r_s}{r}\right)}.$$
If $r = r_s$, then $\mathrm{d}\tau/\mathrm{d}t =0$ and then we have an event horizon. If we take the author at face value and assume the Schwarzschild radius of the universe is equal to the radius of the universe (a strange coincidence?), does this imply we are inside a black hole? The answer is no because the above metric is for a fluid and as the fluid is compressed, its pressure increases. This pressure contributes to the gravitational field, and this invalidates that particular metric as a model of our universe. The universe is better modelled as a pressure-less dust solution (where the stars, planets and galaxies are treated as dust particles). The FLWR metric is such a solution:
$$-c^2 \mathrm{d}\tau^2 = -c^2 \mathrm{d}t^2 + a(t)^2 \frac{\mathrm{d}r^2}{1-kr^2} + r^2 \mathrm{d}\Omega^2,$$
where $k$ has one of the values of the set $\{−1, 0, +1\}$ (for negative, zero, and positive curvature respectively). For radial motion $\mathrm{d}\omega =0$ and for a flat universe, we can set $k=0$ and, after a bit of algebraic manipulation, we obtain:
$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{1 - a(t)^2 \frac{\mathrm{d}r^2}{\mathrm{d}t^2 c^2}},$$
where $a(t)$ is the "scale factor" of the universe, a function of time. For the current epoch, we can set $a(t) =1$ and if we replace $\mathrm{d}r/\mathrm{d}t$ with $v$ in the above equation, it looks identical to the time dilation factor of special relativity. But for the FLWR metric, we have to remember that a distant galaxy on the edge of the universe that appears to be receding away from us at close to the speed can be at rest with the Hubble flow, so we have to set $\mathrm{d}r/\mathrm{d}t=0$ and unlike the event horizon of a black hole, $\mathrm{d}\tau/\mathrm{d}t$ does not go to zero. So, even if the radius of the universe is the same as its Schwarzschild radius, it does not imply we are inside a black hole with an event horizon at the visible radius of the universe when we use the appropriate metric.
Is this simply because to get into the non-observable portion of the
universe, you have to go faster than the speed of light?
This is an intriguing observation. In cosmology, it is perfectly acceptable for objects to be receding from us at greater than the speed of light, and this does not violate relativity as long as locally you are not exceeding the speed of light relative to the local CMB rest frame. Even if a distant galaxy is receding from us at $2c$, it is still possible to move at $+c$ or $-c$ relative to that local galaxy at rest in the Hubble flow. An observer on a galaxy at the visible edge of the universe has no trouble moving outwards to a galaxy beyond our visible range, so she is not in any sense trapped inside the sphere we can see. In fact, that distant observer will see the Earth as being on the edge of their visible sphere.
P.S. If we consider a ballistic model of the universe where galaxies are flying apart in flat Minkowski space simply due to their initial momentum, then the furthest objects would really be moving away from us at close to the speed of light. They would have a very high redshift (high $Z$) due to the relativistic Doppler effect, which is what we observe. In such a model, it would indeed be impossible to escape the visible sphere because doing so would require travelling at greater than $c$ relative to the Earth, which is not allowed in this particular model. However, this is not the generally accepted model.
