Finding a contradiction for analysis problem 
Let $f: \mathbb{N} \to \mathbb{N}$ be a non-decreasing function such that $f(n) \vert n$ is true for only finitely many $n$. Show that there exists a constant $c > 0$ such that $f(n)>cn$ for all $n.$

So trying to find a contradiction one would want to show that $\frac{n}{f(n)} = k$, where $k \in \mathbb{Z^+}$ holds for infinitely many $n$.
From here one can let $g(n) = n - kf(n).$ The problem now becomes to find if $g(n)$ can equal $0$.
My question is that can I use continuity here in order to show this since the problem statement doesn't tell anything about if $f(n)$ is continuous?
One could argue that $g(1) = 1-kf(1) \leqslant 1-k \leqslant 0$ and if one would choose $c=\frac1k$ that would imply that $f(n) > \frac1k n$.
This would result in $g(n) = n-kf(n) > n-k \frac{n}{k} = n-n = 0$
and from Bolzano's theorem the result would follow.

Continuity doesn't really mean much for a function on a discrete set.
We can use induction.
Special case:  Suppose that $f(n)$ never divides $n$.  Then we contend that $f(n)>n$ for all $n$.
Pf:  By induction.  Indeed, $f(1)>1$.  Suppose that we have shown that $f(i)>i$ for all $i<n$.  Then, in particular, $f(n-1)>n-1$. Since $f(n)≥f(n-1)$ we must have $f(n)≥n$.  If we had equality then $f(n)=n$ would divide $n$, contrary to the assumption on $f$, and we are done.
General case:  Since there are only finitely many $n$ for which $f(n)\,|\,n$, we let $K$ be the greatest integer for which $f(K)\,|\,K$.  Then $F(n)=(K+1)f(n)$ can never divide $n$, so we can apply the special case to $F(n)$.  It follows that $$(K+1)f(n)>n\implies f(n)>\frac 1{K+1}n$$ and we are done.
