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.
Note that we say "a neighborhood". This doesn't mean all neighborhoods, otherwise we'd be implying that any function continuous somewhere is bounded everywhere, which is clearly wrong. Consider $1/x$: this function is continuous at, say, $c = 1$. The function is bounded in any neighborhood that doesn't contain 0, so it is bounded for any neighborhood of $c$ smaller than $1$ when $c = 1$.
Now that we have this lemma we can go on to prove that the product of two functions $f$ and $g$ continuous at some point $c$ in their domains is also continuous at $c$.
Proof: Since $f$ is continuous at $c$, there exists some $\delta_1 > 0$, and $M_1 \in \mathbb{R}$ such that if $|x - c| < \delta_1$, $|f(x)| < M_1$ (by our lemma above). Similarly there exist $\delta_2$ and $M_2$ for which the same principles apply for $g$. Now we take $\delta = \inf{\delta_1, \delta_2}$, and $M = \sup{M_1, M_2}, hence $|f(x)| < M$ and $|g(x) < M$. Since $f$ is continuous at $c$, for a given $\epsilon > 0$ there exists $\some $\delta_3 > 0$ such that if $|x - c| < \delta_3$, then $|f(x) - f(c)| < \epsilon / (2(M + 1))$. Once again by similar reasoning for $g$ we can infer the existence of $\delta_4$.
Now if $|x - c| < \delta_5 = \inf{\delta, \delta_3, \delta_4}$, we can perform the following trick:
$$ |f(x)g(x) - f(c)g(c)| = |f(x)g(x) + (f(x)g(c) - f(x)g(c)) - f(c)g(c)| \\
\leq |f(x)||g(x) - g(c)| + |g(c)||f(x) - f(c)| \\
< M|g(x) - g(c)| + M|f(x) - f(c)| \\
< M \cdot \frac{M}{2(M+1)}\epsilon + M \cdot \frac{M}{2(M+1)}\epsilon
= \frac{M}{M+1}\epsilon < \epsilon
$$
And we're done!
