I tried proving inductively but I didn't really go anywhere. So I tried:

Let $3|n(n+1)(n+2)$.

Then $3|n^3 + 3n^2 + 2n \Longrightarrow 3|(n(n(n+3)) + 2)$

But then?

Here is a proof by induction.

The base case $n=0$ or $n=1$ is trivial.

Suppose that $3$ divides $n(n+1)(n+2)$ and consider $(n+1)(n+2)(n+3)$. Expand on $n+3$ and get $(n+1)(n+2)(n+3) = n(n+1)(n+2)+3(n+1)(n+2)$. Since by hypothesis $3$ divides the first term and it clearly divides the second one, it must divide the sum.

This proof can be generalized to $k \mid n(n+1)\cdots(n+k-1)$ using the same technique.

you can write $n=3\cdot k + r$ for some $k\in \mathbb{N}$ and $r\in \{0,1,2\}$. Then exactly one of $n,n+1,n+2$ is divisible by 3.

To do it inductively, we also need a base case: $0 \times 1 \times 2=0$ is divisible by $3$. Then, if $3$ divides $k(k+1)(k+2)=k^3+3k^2+2k$ then \begin{align*} (k+1)(k+2)(k+3) &= k^3+6k^2+11k+6 \\ &= (k^3+3k^2+2k)+3(k^2+3k+2). \end{align*} We know $k^3+3k^2+2k$ is divisible by $3$, by the inductive hypothesis, and $3(k^2+3k+2)$ is divisible by $3$ since $k^2+3k+2$ is an integer. Therefore, by induction, $k(k+1)(k+2)$ is divisible by $3$ for all $k \geq 0$.

Note, if we want to prove it for negative $k$ too, we'd need to perform "induction in reverse".

The binomial coefficients are all integers (various proofs). Hence $$\binom r3=\frac {r(r-1)(r-2)}{6}$$ is an integer. Let $r=n+2$, then $6$ is a factor of your product and $3|6$.

In a similar way you can prove that the product of $k$ consecutive integers is divisible by $k!$ - this cam also be done by induction (which is hidden in the statement "the binomial coefficients are all integers" and implicit in the construction of Pascal's Triangle.

Assume $n(n+1)(n+2) = 3p$. $$(n+1)(n+2)(n+3) = \frac{n(n+1)(n+2)(n+3)}{n} =3p\cdot\frac{n+3}{n}=3p+3\cdot\frac{3p}{n}$$

Notice $\frac{3p}{n}$ is also an integer by assumption.

With this induction step, the base case has to be $n=1$. Prove the case for $n=0$ separately.

Here's a somewhat tedious (but fairly elementary) way. By the division algorithm, we can write $n=3k+r$ for some integers $k,r$ with $0\le r\le 2,$ whence $$\begin{align}n(n+1)(n+2) &= (3k+r)(3k+r+1)(3k+r+2)\\ &= (3k+r)(9k^2+6kr+9k+r^2+3r+2)\\ &= 27k^3+27k^2r+27k^2+9kr^2+18kr+6k+r^3+3r^2+2r\\ &= 3(9k^3+9k^2r+9k^2+3kr^2+6kr+2k+r^2)+r^3+2r.\end{align}$$ All that's left is to show that $r^3+2r$ is a multiple of $3$ for $r=0,1,2$.

Consider the binomial $(x+1)^{n+2}$. The coefficient of the $x^3$ term is

$${n+2\choose 3}={(n+2)!\over 3!(n-1)!}={n(n+1)(n+2)\over 6}$$

Every coefficient of $(x+1)^n$ is an integer for $n$ an integer, therefore $6|n(n+1)(n+2)$ and thus $3|n(n+1)(n+2)$.

Note that this mechanism can apply to any integer, including showing that $1000|n(n+1)(n+2)\cdots(n+998)(n+999)$.

As noted elsewhere, this property of the binomial coefficients is provable by induction, which demonstrates that an inductive proof is truly the proper way to show the question's property. In binomial terms, the inductive step occurs by noting that

$${n\choose k-1}+{n\choose k}={n+1\choose k}$$