Hint $\rm\,\ m,n\mid x\!\iff\! mn\mid mx,nx\!\iff\! mn\mid\overbrace{(mx,nx)}^{\textstyle (m,n)x}\!$ $\rm\iff\! \overbrace{mn/(m,n)}^{\textstyle \ell}\mid x$
Let $\rm\:x = \ell\:$ in $\,(\Leftarrow)\,$ to get $\rm\:m,n\mid \ell,\:$ i.e. $\rm\:\ell\:$ is a common multiple of $\rm\:m,n,\:$ necessarily the least common multiple, since $(\Rightarrow)$ shows $\rm\:m,n\mid x\:\Rightarrow\:\ell\mid x\:\Rightarrow\:\ell\le x.$
Remark $\ $ That $\rm\: m,n\mid x\iff \ell\mid x\,$ is a definition of $\rm\,{\rm lcm}(m,n)\,$ in more general rings since - as above - it implies that $\,\ell\,$ is a common multiple of $\rm\,a,b\,$ that is divisibly least, i.e. it divides every common multiple. See this answer for this universal approach to LCMs and GCDs.
In my experience, in recent number theoretical literature that employs abbreviated notation for the gcd and lcm, the notation $\,(a,b) := \gcd(a,b)\,$ is almost universal, and $\,[a,b] := {\rm lcm}(a,b)\,$ is the most common lcm abbreviation, but far less universal than the gcd notation (so define it if you use it).
One of the primary reasons for adopting the notation $\,(a,b)\,$ for the gcd is that it serves to strongly emphasize the analogy between gcds and ideals that holds in many familiar domains, e.g. in a PID we have $\,(a,b) = (c)\,$ iff $\,c\,$ is a gcd of $\,a,b.\,$ So we can view $\,(a,b)\,$ as either an ideal or a gcd (defined up to a unit factor), and this allows one to give proofs that work both for ideals and gcds, since both satisfy common laws, e.g. associative, commutative, distributive, and $\,(a,b) = (a)\,$ if $\,a\mid b.\,$ For example, see this answer and see this proof of the Freshman's Dream $\,(a,b)^n = (a^n,b^n),\,$ which works for both gcds and invertible ideals.
Remark $\ $ The abbreviated notations chosen in much older textboks were often constrained by the typesetting technology available at the time. Nowadays, no such constraints exist (e.g. using $\TeX).$ Indeed, now one can design new symbols for such purposes that avoid any possibility of confusion with existing notation. While some authors have done just that, none of these notations have yet percolated into the mainstream. That may occur someday if a good design is employed in an influential publication.
Best Answer
Multiples of $a$ are $$ a, 2a, 3a, 4a, \ldots\,. $$ Multiples of $b$ are $$ b, 2b, 3b, 4b, \ldots\,. $$ First we prove existence of a common multiple: $ab$ is a common multiple: it is the $b$th element of the first list and the $a$th element of the second.
It follows that the set of all common multiples is non-empty.
The well-ordering principle says every non-empty set of positive integers has a least member.
Now we prove uniqueness: There cannot be more than one least member since the integers are linearly ordered.
Q.E.D.
In some cases $ab$ is the least one; in other cases there are smaller common multiples than $ab$.
Pedagogical note:
Show a class of students this:
Multiples of $63$: ${}\quad63,126,189,252,315,378,\ldots$
Multuples of $77$: ${}\quad77,154,231,308,385,\ldots$
Then ask them to vote "yes" or "no" on this question: Is it possible that for either this pair of numbers or some other, the two lists go on forever without having any numbers in common? Have them write their votes on paper and turn them in anonymously and see what you get.
PS: OK, it said divisibly least, so let's add something: All multiples of the one that is numerically least -- call it $c$ -- are common multiples of $a$ and $b$ (proving that part is trivial), and here is the "hard" part: There are no others. For, suppose there were another, and call it $d$. Then the remainder on division of $d$ by $c$ would not be $0$ (since $d$ was assumed not a multiple of $c$) and would be less than $c$. And that remainder would be a common multiple of $a$ and $b$. So we have a common multiple of $a$ and $b$ that is less than $c$, thus a contradiction.
H/T @Bill Dubuque