Let $f : \Bbb{R}^n \to \Bbb{R}$ and define
$$
G := \{(x,f(x)) \mid x \in \Bbb{R}^n\},
$$
the graph of $f$. Observe that $G$ is closed, hence measurable.
By the Fubini-Tonelli theorem, applied to the characteristic function/indicator function $\chi_G$ of $G$, we see
$$
\lambda_{n+1}(G) = \int_{\Bbb{R}^{n+1}} \chi_G (x,y) d(x,y) = \int_{\Bbb{R}^n} \int_\Bbb{R} \chi_G (x,y) \, dy \, dx = \int_{\Bbb{R}^n} 0 \, dx = 0,
$$
where the step before the last used that $\chi_G (x,y) \neq 0$ implies $(x,y) \in G$ and hence $y = f(x)$, so that $\chi_G (x,y) = 0$ for all $y \neq f(x)$ and hence in particular for almost all $y \in \Bbb{R}$, so that $\int_\Bbb{R} \chi_G (x,y) dy = 0$.
EDIT: Note that we did use the continuity of $f$ only to conclude that $G$ is indeed measurable. This is the case as soon as $f$ is (Borel/Lebesgue) measurable.
This isn't really relevant to this specific problem, but I just want to add that this result is still true for any Borel-measurable function. This is realized as a trivial consequence of Fubini's Theorem.
Let $m_1$ and $m_2$ denote 1- and 2-dimensional Lebesgue measure, respectively, and let $f: \mathbb{R} \to \mathbb{R}$ be measurable. Then $$ m_2(G_f) = \int_{\mathbb{R}^2} 1_{G_f} dm_2 = \int_{\mathbb{R}} \int_{\mathbb{R}} 1_{G_f}(x,y) dy dx = \int_{\mathbb{R}} m_1(\{ f(x) \}) dx = \int_{\mathbb{R}} 0 dx = 0$$
Best Answer
First assume that $A$ is compact; then $f$ is uniformly continuous on $A$.
Hence fix an $\epsilon$ and pick a $\epsilon_1$ to be decided later so that for some $\epsilon_2$, we have that any $|x-y| < \delta$ implies that $|f(x)-f(y)|< \epsilon$.
Now, note that the measure of the graph of $f$, denoted by $|\Gamma(f)|$, has bound $$|\Gamma(f)| \leq 2 \epsilon |B(0, \delta)| N(\delta)$$ Where $N(\delta)$ denotes the number of balls with radius $\delta$ it takes to cover $A$ and $|B(0,\delta)|$ is the measure of the ball of radius $\delta$ in n dimensions.
Recall that $|B(0,\delta)| \leq C \delta^n$. Also, if $A$ has side lengths $l_i$ in dimension $i$, then $$N(\delta) \leq C \prod_{i=1}^n \frac{l_i}{\delta}$$ (I threw in the constant because I may have been a little sloppy with that bound) Thus $$|\Gamma(f)| \leq 2 K \epsilon$$ for some constant $K$. But $\epsilon$ was arbitrary, hence the result.
For general $A$, since $|\Gamma(f)| = 0$ on every compact $A_n$, $|\Gamma(f)| = 0$.