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 $A_n$, $|\Gamma(f)| = 0$.
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.
Best Answer
It is sufficient to show that the set $G'=\{(x,f(x)) | x \in [0,1)^n \}$ has measure zero.
Let $\epsilon>0$, then since $f$ is uniformly continuous on the compact set $[0,1]^n$, there is some $\delta>0$ such that if $\forall x,x' \in [0,1): \|x-x'\|_\infty < \delta$ (note convenient choice of norm) then $|f(x)-f(x')| < \epsilon$.
Now choose $n$ such that ${ 1\over n} < \delta$ and, with $k = (k_1,...,k_n)$, let $x_k = {k \over n}$ and $R_k = \{x_k\} + [0,{1 \over n})^n$, where each of the $k_i$ range through $0,...,n-1$. Note that $\sum_k m R_k = m [0,1)^n = 1$.
Note that for $x \in R_k$, we have $|f(x)-f(x_k)| < \epsilon$, hence $\{ (x,f(x)) \}_{x \in R_k} \subset R_k \times [f(x_k)-\epsilon, f(x_k) + \epsilon]$ and so $m \{ (x,f(x)) \}_{x \in R_k} \le m R_k \cdot m [f(x_k)-\epsilon, f(x_k) + \epsilon] = 2 \epsilon \, m R_k$.
Hence $m G' \le 2 \epsilon \sum_k \, m R_k = 2 \epsilon$.
Since $\epsilon>0$ was arbitrary, we have $m G' = 0$.