How can I argue that for a number to be divisible by 144 it has to be divisible by 36?

Yes, it's the same. $A\implies B$ is equivalent to "if we have $A$, we must have $B$".

And your proof looks fine. Good job.

If I were to offer some constructive criticism, it would be of the general kind: In number theory, even though we call the property "divisible", we usually avoid division whenever possible. Of the four basic arithmetic operations it is the only one which makes integers into non-integers. And number theory is all about integers.

Therefore, "$n$ is divisible by $144$", or "$144$ divides $n$" as it's also called, is defined a bit backwards:

There is an integer $k$ such that $n=144k$

(This is defined for any number in place of $144$, except $0$.)

Using that definition, your proof becomes something like this:

If $n$ is divisible by $144$, then there is an integer $k$ such that $n=144k$. This gives $$ n=144k=(36\cdot4)k=36(4k) $$ Since $4k$ is an integer, this means $n$ is also divisible by $36$.

Yes that's correct or simply note that

$$n=144\cdot k= 36\cdot (4\cdot k)$$

but $n=36$ is not divisible by $144$.

The $\implies$ symbol is defined as follows:

If $p \implies q$ then if $p$ is true, then $q$ must also be true. So when you say $144 \mid n \implies 36 \mid n$ it's the same thing as saying that if $144 \mid n$, then it must also be true that $36 \mid n$.