The three triangle inequalities are
\begin{align}
x + y &> 1-x-y \\
x + (1-x-y) &> y \\
y + (1-x-y) &> x \\
\end{align}
Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution.
(Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is
$$\int_0^{1/2}\frac1{2-2x}-\frac{2x-1}{2x-2}\,dx=\ln 2-\frac12\approx0.19314$$
I believe that's the answer you calculated.
The lengths of the broken stick pieces $(Y_1, Y_2, Y_3, Y_4)$ are spacings of ordered uniform. The joint distribution is equivalent to
$$ \left(\frac {X_1} {\sum_{i=1}^4 X_i},
\frac {X_2} {\sum_{i=1}^4 X_i},
\frac {X_3} {\sum_{i=1}^4 X_i},
\frac {X_4} {\sum_{i=1}^4 X_i}\right)$$
where $X_i$ are iid exponential random variables. If we ordered the above spacings, the distribution of $(Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)})$ is equivalent to
$$ \left(\frac {X_1/4} {\sum_{i=1}^4 X_i},
\frac {X_1/4 + X_2/3} {\sum_{i=1}^4 X_i},
\frac {X_1/4 + X_2/3 + X_3/2 } {\sum_{i=1}^4 X_i},
\frac {X_1/4 + X_2/3 + X_3/2 + X_4} {\sum_{i=1}^4 X_i}\right)$$
Any $3$ of the $4$ pieces cannot form a triangle if and only if
$$ \begin{cases}
Y_{(1)} + Y_{(2)} < Y_{(3)} \\
Y_{(1)} + Y_{(2)} < Y_{(4)} \\
Y_{(1)} + Y_{(3)} < Y_{(4)} \\
Y_{(2)} + Y_{(3)} < Y_{(4)}
\end{cases}$$
Note that the second inequality is implied by the first, and the third inequality is implied by the fourth. So the probability of no triangle being formed is
$$ \begin{align}
&\Pr\{ Y_{(1)} + Y_{(2)} < Y_{(3)}, Y_{(2)} + Y_{(3)} < Y_{(4)}\} \\
=& \Pr\Bigg\{\frac {X_1} {4} + \frac {X_1} {4} + \frac {X_2} {3} <
\frac {X_1} {4} + \frac {X_2} {3} + \frac {X_3} {2}, \\
& \frac {X_1} {4} + \frac {X_2} {3} + \frac {X_1} {4} + \frac {X_2} {3}
+ \frac {X_3} {2} <
\frac {X_1} {4} + \frac {X_2} {3} + \frac {X_3} {2} + X_4\Bigg\} \\
=& \Pr\{X_1 < 2X_3, 3X_1 + 4X_2 < 12X_4\} \\
=& \int_0^{\infty} \Pr\{2X_3 > x\}\Pr\{12X_4 - 4X_2 > 3x\}e^{-x}dx
\end{align}$$
Note that
$$ \begin{align}
&\Pr\{12X_4 - 4X_2 > 3x\} \\
=& \int_0^{\infty}\Pr\{12X_4 - 4u > 3x\}e^{-u}du \\
=& \int_0^{\infty} e^{-(3x+4u)/12} e^{-u}du \\
=& e^{-x/4}\int_0^{\infty} e^{-4u/3}du \\
=& \frac {3} {4}e^{-x/4}
\end{align}
$$
So the integral become
$$ \int_0^{\infty} e^{-x/2}\frac {3} {4} e^{-x/4}e^{-x}dx
= \frac {3} {4} \int_0^{\infty} e^{-7x/4} dx = \frac {3} {4} \times \frac {4} {7} = \frac {3} {7} $$
This is not a very elegant way but at least it is doable. Looking forward to someone to post a better solution.
Best Answer
Earlier , when I saw this Question , I got the Answer $0.5$ but was hesitant to Post it , because the Simulation by OP was getting $0.45$ limit. With user "Jeroen van der Meer" & user "Dstarred" confirming $0.5$ limit , I think I can safely Post my Analysis.
(1) Shortest 3 Consecutive Pieces :
Let the Stick be broken & the Pieces arranged by Size.
Take the Shortest 3 Consecutive Pieces $(X,Y,Z)$ where $X \le Y \le Z$.
We want $X+Y \ge Z$
We can scale these ( by Dividing throughout by $Z$ ) to get $(x=X/Z,y=Y/Z,z=Z/Z=1)$
What we now want is :
$x+y \ge 1 \tag{C1}$
where
$x \le y \le 1 \tag{C2}$
[[
OBSERVATION 1 : In Case , we have the additional Criteria :
$X+Y+Z = Constant \tag{??C3??}$
then it is equivalent to this Case , where Probability is $1/4$ , though that Criteria is not valid here in the limiting Case.
]]
That Shaded Area ( larger triangle ) is satisfying (2) , while that Highlighted Area ( Smaller triangle ) satisfies (1) , hence that will give the necessary Probability $P=0.25/0.5=0.5$ , matching the Simulations.
[[
OBSERVATION 2 : Simulation Probability is moving from the Initial Case where we have the additional Criteria (C3) to the limiting Case where that Criteria (C3) is irrelevant. The Simulation is moving from $0.25$ ( $n$ close to $3$ ) to $0.5$ ( $n$ very large ) gradually.
]]
(2) Some 3 Consecutive Pieces , not necessarily the Shortest , not necessarily the longest :
Let the Pieces be $P_1,\cdots,P_n$ arranged by Size.
When $(P_1,P_2,P_3)$ can give the triangle , Probability is $P=1/2$.
When $(P_1,P_2,P_3)$ can not give the triangle , it is $(1-P)=1/2$ & we can check $(P_2,P_3,P_4)$ , $(P_3,P_4,P_5)$ , $(P_4,P_5,P_6)$ , $(P_5,P_6,P_7)$ , ETC.
Thus total Probability is $P + P \times (1-P) + P \times (1-P)^2+P \times (1-P)^3\cdots=P/(1-(1-P))=1$
Amazingly , we can guarantee a triangle somewhere among 3 Consecutive Pieces in the limiting Case , no matter what $P$ is !
In other words : (A) If we want acute-angled triangle with some non-Zero Probability , then we can guarantee that among 3 Consecutive Pieces when $n$ is very large. (B) If we want obtuse-angled triangle with some non-Zero Probability , then we can guarantee that among 3 Consecutive Pieces when $n$ is very large. (C) If we want all irrational sides of the triangle with some non-Zero Probability , then we can guarantee that among 3 Consecutive Pieces when $n$ is very large. Necessary Criteria is that $P$ must be non-Zero.