Proposition: If $f : A \to \mathbb{R}$ is continuous at $c \in A$, $kf$ is continuous at $c$ for $k \in \mathbb{R}$
Proof: Let $\epsilon > 0$ be given. Since $f$ is continuous at $c$, $\exists \delta > 0$ such that if $|x - c| < \delta$ then $|f(x) - f(c)| < \epsilon / (|k| + 1)$. Thus if $|x - c| < \delta$:
$$ |kf(x) - kf(c)| = |k||f(x) - f(c)| \leq \frac{|k|\epsilon}{|k| + 1} = \frac{|k|}{|k|+1}\epsilon < \epsilon $$
If we want to show that the product of two continuous functions is continuous, we need the following lemma:
Lemma: If $f : A \to \mathbb{R}$ is continuous at $c \in A$, $f$ is bounded in a neighborhood of $c$.
Proof: Let $\epsilon = 1$. As $f$ is continuous at $c$ we know there exists some $\delta > 0$ such that if $|x - c| < \delta$, then $|f(x) _ f(c)| < \epsilon = 1$. By the triangle inequality, we can go on to say that $|f(x)| - |f(c)| \leq |f(x) - f(c)| < 1$, therefore $|f(x)| < 1 + |f(c)| = M$. Thus if $|x - c| < \delta$, then $|f(x)| < M$, i.e. it is bounded.