Choose two points $A$ and $B$, randomly with uniform distribution, along a line segment ($I$ of length $1$, say). What is the probability that the three demarcated subsegments satisfy the triangle inequality?
The answer is $1/4$, as proven elsewhere on this site and elsewhere.
Here's a "proof" that the answer is $1/2$:
Choose $A$ first. (Ignore the $0$-probability case that it's an endpoint or the midpoint of $I$.) Wherever $A$ is, $B$ must be within distance $1/2$ of $A$ on $A$'s longer side, of which the probability is $1/2$.
Obviously that "proof" isn't a proof, since the result is false. But what's wrong with it?
Let's place $A$ for the moment at the point $1/3$ on the interval $[0,1]$.
While it is true that placing $B$ anywhere from $1/3$ to $5/6$ means it is close enough to $A$ that $AB$ has length $\ell<1/2$, the area from $1/3$ to $1/2$ still has the problem of the segment from $B$ to the far end being longer than $1/2$.
So in short you've only considered two of the pieces that have to be of length $\ell<1/2$: the other condition, that $A$ and $B$ are on opposite sides of the point $1/2$, is not yet met.