Assertion:

There exists a real number x such that x² + 1 ≤ 0.

Disproof:

Let x be an arbitrary real number.

The given inequality x² + 1 ≤ 0 can be rewritten as x² ≤ −1.

Since x² ≥ 0 holds for every real number x, this is a contradiction

Thus, there does not exist a real number x such that x² + 1 ≤ 0.

Therefore, the given assertion is false.