It's possible to construct silly confidence intervals that are technically valid, i.e. have the claimed coverage, but it's not obligatory. So I don't know why doing so should demonstrate a flaw in "confidence interval methodology" (whatever that is)—you could have constructed a more appropriate one for your purposes.
A more orthodox approach to this particular problem:
First, note that you can reëxpress the sufficient statistic $(Y,Z)$ to contain an obvious ancillary component: the sample range $Z-Y$. It's a very strong precision index: as it approaches $a$, $\theta$ is known almost exactly. So any inference should be conditional on its observed value $z-y$.
Second, the sample minimum $Y$ conditional on the observed value of the range is uniformly distributed between $\theta$ and $\theta+a-r$. Confidence intervals based on this distribution could be constructed & would be well behaved. Still, the likelihood for $\theta$ is flat between $z-a$ and $y$, so it wouldn't be possible to say for any confidence interval with less than 100% coverage that it contained values of $\theta$ less discrepant with the data than those outside it. Best to stick with the 100% interval.
[Response to comments:
(1) Your confidence interval does have some undesirable properties* but that's no reason to tar all confidence intervals with the same brush. You only took unconditional coverage into account when you derived it & therefore have no right to complain that unconditional coverage is all it gives you.
(2) You're right that $\left(Y-\left(1-\frac{1-\gamma} 2\right)(a-r),Y-\frac{1-\gamma} 2(a-r)\right)$ is a valid C.I., conditional on $r$, but why not use $(Y-\gamma(a-r),Y)$? The likelihood is $\frac{1}{a-r}$ for all values of $\theta$ between $y+r-a$ and $y$, & I don't see any generally compelling reason to honour an arbitrary 95% of those values with inclusion in my confidence interval. The 100% interval $(y+r-a,y)$ is preferable as it separates zero-likelihood values of $\theta$ from those with positive likelihood.
*At least for inference on $\theta$ from a single sample: there may be some applications—say in quality control, where consideration of coverage over repeated samples is more than a thought experiment—for which it could have some use.]
Best Answer
I am wondering how you get this confidence interval. My method could be using the pivot, but result in different interval.
Let $Q = \frac{X_{(n)}}{\theta}$, where $X_{(n)}$ is the largest order statistics of $X_1, ......, X_n$. Then,
$$P(Q \le t) = \prod_{i=1}^nP(X_i \le t\theta) = t^n$$
, so Q is a pivot (independent of the parameter $\theta$). Take $c_n = \alpha^{\frac{1}{n}}$ we obtain $P(Q \le c_n) = \alpha$ and given that $P(Q \le 1) = 1$, we have
$$1 - \alpha = P(c_n \le Q \le 1) = P(c_n \le \frac{X_{(n)}}{\theta} \le 1) =P(X_{(n)} \le \theta \le \frac{X_{(n)}}{c_n})$$
So the $1-\alpha$ confidence interval for $\theta$ is $\Big(X_{(n)}, \frac{X_{(n)}}{\alpha^{\frac{1}{n}}}\Big)$
Finally you can plug in n = 2 and n = 100 to get different answers. (But sorry it is not the same you give here.)