Prevent bash from adding single quotes to variable

Problem: I'm writing a script that performs several HTTP requests with curl and I want to add the headers to a variable, CURL_HEADERS, so that I don't have to type them out constantly. When I echo the CURL_HEADERS variable in the curl command, single quotations appear where I don't want them. How can I prevent this? (The code below is simplified for the sake of clarity)

Code

#!/usr/bin/env bash
AUTH_KEY='1234'
set -x
CURL_HEADERS='-H "Authorization: Basic '${AUTH_KEY}'" -H "Content-Type: application/json"'
echo "${CURL_HEADERS}"
curl -s $(echo "${CURL_HEADERS}") 'http://www.example.org' > /dev/null
set +x

Expected Output:

+ CURL_HEADERS='-H "Authorization: Basic 1234" -H "Content-Type: application/json"'
+ echo '-H "Authorization: Basic 1234" -H "Content-Type: application/json"'
-H "Authorization: Basic 1234" -H "Content-Type: application/json"
++ echo '-H "Authorization: Basic 1234" -H "Content-Type: application/json"'
+ curl -s -H "Authorization: Basic 1234" -H "Content-Type: application/json" http://www.example.org
+ set +x

Actual Output

+ CURL_HEADERS='-H "Authorization: Basic 1234" -H "Content-Type: application/json"'
+ echo '-H "Authorization: Basic 1234" -H "Content-Type: application/json"'
-H "Authorization: Basic 1234" -H "Content-Type: application/json"
++ echo '-H "Authorization: Basic 1234" -H "Content-Type: application/json"'
+ curl -s -H '"Authorization:' Basic '1234"' -H '"Content-Type:' 'application/json"' http://www.example.org
+ set +x

A reasonably straightforward solution is to use a bash array to store the four arguments you will want to pass:

CURL_HEADERS=(
             '-H' "Authorization: Basic ${AUTH_KEY}"
             '-H' 'Content-Type: application/json'
)

curl -s "${CURL_HEADERS[@]}" 'http://www.example.org' > /dev/null

Unlike scalar variables, which are just ordinary strings of ordinary characters no matter how many quotes they might contain, arrays are lists of strings, each one distinguished from each other one. In this sense, bash is just like almost every other programming language.

This problem, and the solution I suggest as well as several other ones, is well-described in the Bash FAQ entry 50 (I'm trying to put a command in a variable, but the complex cases always fail!), which is worth reading in detail. (Link taken from a comment by @John1024.)