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.
Not an answer, but it might help to progress further. [The derivation that follows provides a strict -but practically useless- bound. The second part of the answer has some results that might be of interest, but they are merely empirical]
Let's consider the (much more restricted) event that the six lengths form a tetrahedron in whatever order. In the linked pdf this set of lengths is called "completely tetrahedral", and a necessary-sufficient condition is given (Theorem 4.2) which is equivalent to the following: $ u \le \sqrt{2}v $, where $u,v$ are the maximum and minimum lengths. This, informally, would correspond to "almost regular" tetrahedra.
Let's then compute the probability that the lengths are completely tetrahedral. Because the points are chosen at random, uniformly, the lengths are probabilistically equivalent to a set of iid exponential variables with arbitrary parameter, conditioned to a constant sum. Because we are only interested in ratios, we can even ignore this conditioning.
Now, the joint probability of the maximum and minimum of a set of $n$ iid variables is given by
$$ f_{u,v}= n(n-1) f(u) f(v) [F(u) -F(v)]^{n-2}, \hskip{1cm} u\ge v$$
In our case: $n=6$, $f(u)=e^{-u}$, and the probability that $u<a \, v$ is a straightforward integral, which gives:
$$P(u<a \, v)= 5 (a -1)\left( \frac{1}{1+5\,a}-\frac{4}{2+4\,a}+\frac{6}{3+3\,a}-\frac{4}{4+2\,a}+\frac{1}{5+1\,a} \right)$$
And $P(u<\sqrt{2} v) \approx 7.46 \times 10^{-5}$
This should be a strict bound on the desired probability, but, surely, far from tight.
[Update: indeed, the bound is indeed practically useless, it corresponds to an irrelevant tail. The probability, as per my simulations, is around $p=0.06528$ ($N=10^9$ tries, $3 \, \sigma \approx 2.3 \times 10^{-5}$), which agrees with other results.]
The only empirical result that might be of interest: It's easy to see that, from the $6!$ possible permutations of the lenghts, we can restrict ourselves to $30$, from symmetry considerations; now, from my simulations, I've found that it's sufficient to consider 7 permutations, the first two being already enough for more than $90\%$ of the successes;
and the (need to consider the) seventh one is extremely small. These permutations are:
$$p_1 = [0 , 1 , 4 , 5 , 3 , 2] \hskip{1 cm} (0.75837)\\
p_2 = [0 , 1 , 4 , 3 , 5 , 2] \hskip{1 cm} (0.15231)\\
p_3 = [0 , 2 , 4 , 1 , 5 , 3] \hskip{1 cm} (0.08165)\\
p_4 = [0 , 1 , 4 , 5 , 2 , 3] \hskip{1 cm} (0.00404)\\
p_5 = [0 , 1 , 4 , 2 , 5 , 3] \hskip{1 cm} (0.00245)\\
p_6 = [0 , 1 , 3 , 5 , 4 , 2] \hskip{1 cm} (0.00116)\\
p_7 = [0 , 1 , 3 , 4 , 5 , 2] \hskip{1 cm} (0.00002)\\
$$
The length indexes correspond to a sorted array (say, ascending), and following the convention of the linked paper: the first three sides have a common vertex, the following three are the corresponding opposite sides (so, for example, in the first permutation, and by far the most favorable one, the longest and shortest sides are opposite). The numbers on the right are the probability that this permutation (when one tries in the above order) is the successful one (given that they form a tetrahedron). I cannot be totally sure if there is some rare case that requires other permutation (very improbable, I'd say), but I'm quite sure (unless I've made some mistake) that the set cannot be further reduced.
Best Answer
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.