No, this is incorrect. You are thinking of the event horizon as a "place in space". It's not. It's an outgoing light ray. And "locations" with a given radius inside the horizon are not places in space either; they are moments of time.

Here's how to unpack what I just said in more detail. What kind of curve is a curve of constant r (and constant theta, phi if we include the angular coordinates)? Our ordinary intuition about how "space" works says that a curve of constant r is timelike; that is, it is a possible worldline for an object. This means that an object can "stay in the same place" by staying at constant r for all time.

If a black hole is present, however, this ordinary intuition is only valid outside the event horizon. Outside the horizon, curves of constant r are timelike. However, the event horizon is a curve of constant r, and it is not timelike: it's null. It's the path of an outgoing light ray. A light ray always moves at the speed of light; it can never "stay in the same place". So you can't think of the event horizon as "a place in space" the way you can think of a curve of constant r outside the horizon.

Inside the horizon, a curve of constant r is spacelike. The physical meaning of a spacelike curve is that it is the set of all events that happen "at the same time" for some observer. So a curve of constant r inside the horizon is "a moment of time" rather than "a place in space".

So a light ray that is emitted "1 foot below the horizon" is really emitted "1 foot to the future of the horizon". Or, since light travels about 1 foot per nanosecond, the light from your feet is emitted 1 nanosecond to the future of the horizon. That light can't travel "upward" to meet your feet; that would mean traveling into the past. Instead, your feet have to fall through the horizon to meet the light.