How can I get this script to error exit based on result of for loop?

Replace:

done

with:

done || exit 1

This will cause the code to exit if the for loop exits with a non-zero exit code.

As a point of trivia, the 1 in exit 1 is not needed. A plain exit command would exit with the exit status of the last executed command which would be false (code=1) if the download fails. If the download succeeds, the exit code of the loop is the exit code of the echo command. echo normally exits with code=0, signally success. In that case, the || does not trigger and the exit command is not executed.

Lastly, note that set -o errexit can be full of surprises. For a discussion of its pros and cons, see Greg's FAQ #105.

Documentation

From man bash:

for (( expr1 ; expr2 ; expr3 )) ; do list ; done
First, the arithmetic expression expr1 is evaluated according the rules described below under ARITHMETIC EVALUATION. The arithmetic expression expr2 is then evaluated repeatedly until it evaluates to zero. Each time expr2 evaluates to a non-zero value, list is executed and the arithmetic expression expr3 is evaluated. If any expression is omitted, it behaves as if it evaluates to 1. The return value is the exit status of the last command in list that is executed, or false if any of the expressions is invalid. [Emphasis added]

If you prefer to stop the loop on the first failure, and also to avoid set -e, you can do something like this:

for i in `seq 1 10`; do
  run_command || exit 1;
done

If you have errexit set, then the false statement should cause the script to exit immediately. Same thing if the curl command failed.

You're example script, as written, should exit after the first curl command failure the first time it calls false if errexit is set.

To see how it works (I use the shorthand -e for to set errexit:

$ ( set -e;  false; echo still here )
$

$ ( set +e;  false; echo still here )
still here
$

So if the curl command executes more than once, this script doesn't have errexit set.