My attempt:
Suppose $[a, b] = t =$ lcm of $a, b.$
By definition of lcm $a, b \mid t$.
If $a, b \mid t$ and $a, b \mid c$, then $|t| \le |c|$ since $t$ is the smallest such integer. So, $t \mid c$.
Assume $a, b \mid t$. It is in the definition of lcm, so it passes.
Also, assume $a, b \mid c \to t \mid c$. Then $t \mid c \to |t| \le |c|$ and since $t > 0, t \le c$.
Please, check and see.
You wrote "iff $a, b \mid t$ and $a, b \mid c \to t \mid c$". I think this was to be parsed as
"$\text{iff }\Big( \Big(a, b \mid t\text{ and }a, b \mid c\Big) \to t \mid c\Big)$",
not as
"$\text{iff }\Big( (a, b \mid t)\text{ and }(a, b \mid c \to t \mid c)\Big)$".
You have $a,b\mid t$ by definition and $a,b\mid c$ by hypothesis. Is $\gcd(t,c)$ equal to $t$ or less than $t$? Do you know that $\gcd(t,c) = at+bc$ for some integers $a,b$? If not, then maybe that part is most of the work you'll need to do. Can you use that to to show that $a,b\mid \gcd(t,c)$? If $a,b\mid \gcd(t,c)$ and $t$ is the smallest number that $a,b$ both divide, then $\gcd(t,c)$ cannot be smaller than $t$, so it must be equal to $t$. Since $\gcd(t,x)\mid c$, we then get $t\mid c$.