Show that if $d(n)$ is $O(f(n))$, then $ad(n)$ is $O(f(n))$, for any constant $a > 0$?
No, it is not sufficient to say "let $d(n) = n$ which is $O(f(n))$. Therefore $ad(n) = an$ which is trivially $O(f(n))$". Although that is a reasonable way to to understand the proposition quickly, it is neither sufficient nor necessary. It cannot be considered as a proof. It can easily lead to misunderstanding if communicated.

To show that "if $d(n)$ is $O(f(n))$, then $ad(n)$ is $O(f(n))$, for any constant $a &gt; 0$", let us apply the relevant definitions. 

$$\begin{align*}
d(n)\text{ is }O(f(n)) 
&amp;\Longrightarrow \limsup_{n\to\infty}\dfrac{|d(n)|}{f(n)} &lt;\infty\\
\left(\text{since } \limsup_{n\to\infty}\dfrac{a|d(n)|}{f(n)}=a\limsup_{n\to\infty}\dfrac{|d(n)|}{f(n)}\right) &amp;\Longrightarrow\limsup_{n\to\infty}\dfrac{a|d(n)|}{f(n)} &lt;\infty\\
&amp;\Longrightarrow\limsup_{n\to\infty}\dfrac{|ad(n)|}{f(n)} &lt;\infty\\ &amp;\Longrightarrow ad(n)\text{ is }O(f(n)).\\
\end{align*}$$

The proof above is rigorous, although it is hardly the way we as humans understand the proposition. Here is another approach.

$$\begin{align*}
d(n)\text{ is }O(f(n)) 
&amp;\Longrightarrow |d(n)|\text{ is bounded above by } cf(n)\text{ when $n$ is large enough for some constant } c\\
&amp;\Longrightarrow |ad(n)|\text{ is bounded above by } acf(n)\text{ when $n$ is large enough for some constant } c\\
(\text{let } c'=ac)\ \ &amp;\Longrightarrow |ad(n)|\text{ is bounded above by } c'f(n)\text{ when $n$ is large enough for some constant } c'\\
&amp;\Longrightarrow ad(n)\text{ is }O(f(n)).\\
\end{align*}$$

The approach above can be considered as a proof among people who are familiar with the stuff. It is probably the way to understand the proposition as well. You can imagine that the graph of $cf(n)$ lies above the graph of $d(n)$, and, hence, the graph of $acf(n)$ lies above the graph of $ad(n)$.