Sorry, not true. What I was showing was that you could get the standard error for any value of x using fit["SinglePredictionBands"]. (You take the difference in the upper and lower values and divide by twice the t cut-off value.)

fit["SinglePredictionErrors"] only gets you the standard errors associated with the x values in your data. In other words, fit["SinglePredictionErrors"][[3]] gets you the standard error associated with the third element in your data. That just also happens to be x=3 for your data. But in general that certainly won't be the case.

Thank you for your answer, Jim.

But, frankly, I'm not a guru in statistic. I want to understand a simple thing: what value to use for estimate of the standard error of a prediction. Can I state it as follows: I need to use the halfwidth of the confidence interval, rather than the estimate of the standard error, since the former takes into account the finite size of my series, and the latter doesn't?

That's because SinglePredictionErrors gives you the estimates of the standard error and the halfwidth of the confidence interval gives you the t-value times the estimate of the standard error.

fit["SinglePredictionBands"] /. x -> # & /@ data[[All, 1]]
(* {{0.455359, 1.81737}, {1.45006, 2.76207}, {2.43915, 3.71237}, {3.42211, 4.6688}, {4.39854, 5.63176}, 
{5.36824, 6.60146}, {6.3312, 7.57789}, {7.28763, 8.56085}, {8.23793, 9.54994}, {9.18263, 10.5446}}*)

fit["SinglePredictionConfidenceIntervals"]
(* {{0.455359, 1.81737}, {1.45006, 2.76207}, {2.43915, 3.71237}, {3.42211, 4.6688}, {4.39854, 5.63176}, 
{5.36824, 6.60146}, {6.3312, 7.57789}, {7.28763, 8.56085}, {8.23793, 9.54994}, {9.18263, 10.5446}}*)

t = InverseCDF[StudentTDistribution[8], 1 - (1 - CLOneSigma)/2];
fit["PredictedResponse"] + (# {-1, 1} & /@ (t*fit["SinglePredictionErrors"]))
(* {{0.455359, 1.81737}, {1.45006, 2.76207}, {2.43915, 3.71237}, {3.42211, 4.6688}, {4.39854, 5.63176}, 
{5.36824, 6.60146}, {6.3312, 7.57789}, {7.28763, 8.56085}, {8.23793, 9.54994}, {9.18263, 10.5446}}*)

or equivalently

spe1 = fit["SinglePredictionErrors"];
t = InverseCDF[StudentTDistribution[8], 1 - (1 - CLOneSigma)/2];
spe2 = Table[((#[[2]] - #[[1]])/(2 t)) &[fit["SinglePredictionBands"] /. x -> data[[i, 1]]], {i, Length[data]}];
spe2/spe1
(* {1., 1., 1., 1., 1., 1., 1., 1., 1., 1.} *)