Prove that if $a|b$ and $a|c$ then $a|(sb+tc)$ for all $s, t \in \mathbb{Z}$ Would this be the same thing as saying "Prove that if $a|b$ and $a|c$ then $a|(sb+tc)$ for any $s, t \in \mathbb{Z}$"? I can do the proof for any integers $s$ and $t$, but if any and all mean the same thing, then I can do this proof. If not, how can I do this?

In response to your question about any and all, they generally mean the same thing. The symbol $\forall$ can be read "for all" or "for every" or "for any" or "for each". These all mean the same thing.
This is how you could write the proof, with meticulous rigor. 

Let $a,b,c \in \mathbb{Z}$ with $a \mid b$ and $a \mid c$. 
Let $s,t \in \mathbb{Z}$ be arbitrary. 
Then by definition, $\exists\, k_1, k_2 \in \mathbb{Z}$ such that $$b = ak_1$$
$$c = ak_2$$
Then $$sb = s(ak_1)$$
$$tc = t(ak_2)$$
By commutativity and associativity,
$$sb = a(sk_1)$$
$$tc = a(tk_2)$$
Adding these equations, we get
$$sb + tc = a(sk_1) + a(tk_2)$$
By distributivity,
$$sb + tc = a(sk_1 + tk_2)$$
Since $\mathbb{Z}$ is closed under addition and multiplication, $$(sk_1 + tk_2) \in \mathbb{Z}$$
Thus, by definition, $$a \mid (sb+tc)$$
