You don't compare the individual points to conclude a treatment effect. You see whether the lines for the treatment and control are different.
In some circumstances, the fitted lines might be parallel, and just the difference in intercept is of interest. In others, both the intercept and slope might differ, and any difference would be of interest.
Testing point vs line in ordinary regression (not errors-in-variables, which is more complicated):
It's not correct to check if data values for another are in the confidence interval because the data values themselves have noise.
Call the first sample $(\underline{x}_1,\underline{y}_1)$, and the second one $(\underline{x}_2,\underline{y}_2)$. Your model for the first sample is $y_1(i) = \alpha_1 + \beta_1 x_{1,i} + \varepsilon_i$, with the usual iid $N(0,\sigma^2)$ assumption on the errors.
You want to see if a particular point $(x_{2,j},y_{2,j})$ is consistent with the first sample. Equivalently, to check whether an interval for $y_{2,j} - \left(\alpha_1 + \beta_1 x_{2,j}\right)$ includes 0 (notice the points are second-sample, the line is first-sample).
The usual way to obtain such CI would to construct a pivotal quantity, though one could simulate or boostrap as well.
However, since in this illustration we're doing it for a single point, under normal assumptions and with ordinary regression conditions, we can save some effort: this is a solved problem. It corresponds to (assuming sample 1 and sample 2 have a common population variance) checking whether one of the sample 2 observations lies within a prediction interval based on sample 1, rather than a confidence interval.
Of course you're confused. The web page you're reading is confusing. The opening sentence is incorrect and the opening paragraph is riddled with problems primarily focusing on a magnitude of significance. In fact, the only paragraph on the page that seems satisfactory to me is the one from which your quote is taken.
Nevertheless, there is a consistency. If significance were to have an amount it's not the p-value, it's the alpha value. And, the meaning of alpha is explained there well. Therefore, according to the site, one can infer that the amount of significance can only change due to lowering the alpha value before the study is run and the discovered p-values or CI's make absolutely no difference. The latter part of that is correct. But, generally it's advised that there just really isn't such a thing as more or less significant. I don't know anyone personally who would endorse that website's interpretation of the term "more significant".
It appears, from your question, that you want to say one of your tests is more significant than the other. You cannot do that, even following the referenced website, because you calculated 95% CI's for both and therefore used the same alpha values (alpha is used to determine the % of the CI). You can't change alpha afterwards. Furthermore, it's quite unlikely that you even want to say "more significant" for treatment A. You would probably like to say that it had a larger effect. In that case, what you would do would be to compare the effects of treatment A and treatment B directly.
The test comparing treatment A and B would fail to show they were different. That's evident in the large overlap in your confidence intervals. I'm suspecting they're fairly wide CI's as well. I say "suspect" given that the width is substantially greater than the magnitude relative to 0. I'd need to know much more about the data to make a confident qualitative statement that they are wide because it depends on several factors. However, given the apparent width, the best thing you can say is that the CI's of A and B largely overlap and it cannot be determined from these small samples whether the observed differences are simply due to variable samples because those CI's are also rather wide.
Best Answer
Is this a simple regression (one predictor)? If so, there's nothing to do -- if the slope is significantly different from zero, so are two distinct means. The t-statistic for the difference is just the t-statistic for the slope (apart, perhaps, from a sign change). Looking as absolute values:
$$\left|\frac{\hat{y}_1-\hat{y}_2}{\sqrt{\text{Var}(\hat{y}_1-\hat{y}_2)}}\right|=\left|\frac{(x_1-x_2)\hat{\beta}}{\sqrt{\text{Var}[(x_1-x_2)\hat{\beta}}]}\right|$$
$$=\left|\frac{(x_1-x_2)\hat{\beta}}{(x_1-x_2)\sqrt{\text{Var}(\hat{\beta})}}\right|$$
$$=\left|\frac{\hat{\beta}}{\sqrt{\text{Var}(\hat{\beta})}}\right|$$
If you have a multiple regression it's somewhat more complicated, but a similar calculation could be used to build a test for a change in mean.
That's a different thing. For example, it ignores the correlation between the predictions.
Are you really interested in statistical significance or whether there's a practical difference?