A classic sorting code-golf question

Haskell, 87 80 89

s%[]=s
(x:a)%q|x<=q!!0=x:a%q
p%q=q%p
j(x:y:s)=x%y:j s
j a=a
r[x]=x
r s=r$j s
s=r.map(:[])

This is merge sort, implemented from the bottom up. first we package every element into it's own list, and then merge them two-by-two, and again merge them two-by-two, until we're left with one list.

(%) is the merge function
j merges pairs in a list of lists
r merges a complete list of lists
s is the sorting function.

Usage: Run an interpreter, and enter s [3,5,2,6,7].

Edit: the way I was merging things before wasn't the right order, So to fix it I needed 9 more characters.

JavaScript (ES6), 195 193 191 189 188 186 183 182 179 174 172 bytes

This is a heapsort implementation. I expect someone to come up with a shorter mergesort, but I like this one :P

Update: R mergesort beaten. Ruby up next :D

S=l=>{e=l.length
W=(a,b)=>[l[a],l[b]]=[l[b],l[a]]
D=s=>{for(;(c=s*2+1)<e;s=r<s?s:e)s=l[r=s]<l[c]?c:s,W(r,s=++c<e&&l[s]<l[c]?c:s)}
for(s=e>>1;s;)D(--s)
for(;--e;D(0))W(0,e)}

Test (Firefox)

S=l=>{e=l.length
W=(a,b)=>[l[a],l[b]]=[l[b],l[a]]
D=s=>{for(;(c=s*2+1)<e;s=r<s?s:e)s=l[r=s]<l[c]?c:s,W(r,s=++c<e&&l[s]<l[c]?c:s)}
for(s=e>>1;s;)D(--s)
for(;--e;D(0))W(0,e)}

document.querySelector('#d').addEventListener("click",function(){a=JSON.parse(document.querySelector('#a').value);S(a);document.querySelector('#b').innerHTML=JSON.stringify(a)});

<textarea id="a">[9,4,1,2,7,3,5,8,6,10]</textarea><br><button id="d">Sort</button><br><pre id="b"></pre>

Python3, 132 bytes

def S(l):
 if len(l)<2:return l
 a,b,o=S(l[::2]),S(l[1::2]),[]
 while a and b:o.append([a,b][a[-1]<b[-1]].pop())
 return a+b+o[::-1]

Simple mergesort. Lots of bytes were spent to make sure this actually runs in O(n log n), if only the algorithm, but not the implementation needs to be O(n log n), this can be shortened:

Python3, 99 bytes

def m(a,b):
 while a+b:yield[a,b][a<b].pop(0)
S=lambda l:l[1:]and list(m(S(l[::2]),S(l[1::2])))or l

This is not O(n log n) because .pop(0) is O(n), making the merge function O(n^2). But this is fairly artificial, as .pop(0) could easily have been O(1).