Bayesian credible interval. In Bayesian estimation of binomial success probability $\theta,$ suppose we begin with the prior distribution $\theta \sim \mathsf{Beta}(2. 2).$
Upon observing $x = 45$ successes in $n = 100$ trials, we have likelihood proportional to $\theta^{45}(1-\theta)^{55}.$
Then according to Bayes' Theorem, the posterior distribution is
$\theta \sim \mathsf{Beta}(47. 57),$ and 95% Bayesian posterior probability interval
(credible interval) $(0.358, 0.548).$
qbeta(c(.025, .975), 47, 57)
[1] 0.3578055 0.5477928
If we believe the prior and the integrity of the data, then be must believe
the posterior probability distribution is correct. According to the posterior
distribution, $P(0.358 < \theta < 0.548) = 0.95.$
I'm puzzled by your sentence, "[T]hat doesn't guarantee the true parameter is captured by this credible interval." There is no absolute "guarantee" attached to any interval estimate [other than $(0,1)$]. However, under the posterior probability
distribution, there is probability 95% that the credible interval covers $\theta.$
Agresti-Cooll frequntist confidence interval. For the same data as above
we have a 95% A-C confidence interval %(0.356, 0.548).$
p = 47/104; p+qnorm(c(.025,.975))*sqrt(p*(1-p)/104)
[1] 0.3562732 0.5475730
One frequentist interpretation is that the process by which this CI was obtained
would contain the true unknown value of $\theta$ over the long run in
future repetitions of the experiment that produced the data.