List lines beetween 2 keywords using grep/sed/awk - regex

Question

I have a sas log file and I want to list only those lines that are between two words: data and run.
File can contain many such words in many lines, for example:
MPRINT: data xxxxx;
yyyyy
xxxxxx
MPRINT: run;
fffff
yyyyy
data fff;
fffff
run;
I would like to have lines 1-4 and 8-10.
I tried something like
egrep -iz file -e '\sdata\s+\S*\s+(.|\s)*\srun\s' but this expression lists all lines between first begin and last end ((.|\s) is for the purpose of new line character).
I may also want to add additional words to pattern between data and run like:
MPRINT: data xxx;
fffff
NOTE: ffdd
set fff;
xxxxxx
MPRINT: run;
data fff;
yyyyyy
run;
In some cases I would like to list only lines between data and run where there is set word in some line.
I know there are many similar threads, but I didn't find any when keywords can repeat multiple times.
I'm not familiar awk or sed but if it can help I can also use it.
[Edit]
Note that data and run are not necessarily on the beginning of the line (I updated the example). Also there can't be any other data between data and run.
[Edit2]
As Tom noted every line that I was looking for started with MPRINT(...):, so filtered those lines.
Anubhava answer helped me the most with my final solution so I mark it as an answer.
Final expression looked like this :
grep -o path -e 'MPRINT.*' | cut -f '2-' -d ' '|
grep -iozP '(?ms) data [^\(;\s]+.*?(set|infile).*?run[^\n]*\n

Answer 1

You may use this gnu grep command witn -P (PCRE) option:
grep -ozP '(?ms).*?data .*?run[^\n]*\n' file
If you only want to print block with line starting from set then use:
grep -ozP '(?ms).*?data .*?^set.*?run[^\n]*\n' file
MPRINT: data xxxxx;
yyyyy
set fff;
xxxxxx
MLOGIC: run;
You may use this awk to print between 2 keywords that must contain a line starting with set:
awk '/data / {
p=1
}
p && !y {
if (/^set/)
y=1
else
buf = buf $0 ORS
}
y {
if (buf != "")
printf "%s", buf
buf=""
print
}
/run/ {
p=y=0
}' file
MPRINT: data xxxxx;
yyyyy
set fff;
xxxxxx
MLOGIC: run;
If you just want to print data between 2 keywords in awk, it is so simple:
awk '/data /,/run/' file

Answer 2

For what i understand the following will do the trick
sed -n '/data.*;/,/run;/p' $FILENAME
Note that the '.*' after data can be improved by something like [a-z|A-Z]{5} that you protect against matching the word data somewhere in the middle
From there matching from data to set would already require some external decision processes, so the command would be
sed -n '/data.*;/,/set.*;/p' $FILENAME
(Probably learned along the way from How to use sed/grep to extract text between two words?)

Answer 3

Just try (?s)data.+?run;
Explanation:
(?s) - single line mode, . matches newline character
data - match data literally
.+? - match one or more of any character (including neline), non-greedy due to ?
run; - match run; literally
Demo