Prove by induction that $\bigg \vert\prod_{k=1}^{n} a_k - \prod_{k=1}^{n} b_k \bigg \vert \leq \sum_{k=1}^{n} | a_k - b_k|$.

Let $A_{n} = \prod_{k=1}^{n}a_{k}$ and $B_{n} = \prod_{k=1}^{n}b_{k}$.

The base case $n = 1$ obviously holds true.

Suppose we have $|A_{n} - B_{n}| \leq \sum_{k=1}^{n}|a_{k} - b_{k}|$.

Then we have $|A_{n +1} - B_{n+1}| \leq |a_{n+1} - b_{n+1}||A_{n}| + |A_{n} - B_{n}||b_{n+1}| \leq \sum_{k=1}^{n+1} |a_{k} - b_{k}|$ (I have used the triangle inequality, the hypothesis, and the inductive step]. This completes the induction.

A different proof, possibly interesting: The problem said "prove by induction". In some sense any proof of anything for every $n\in\Bbb N$ is based on induction at some point. Here the induction is in the "product rule" for the derivative of the product of $n$ functions:

$\left(\prod_{j=1}^nf_j\right)'=\sum_{k=1}^nf_k'\prod_{j\ne k}f_j$.

Now define $$f(t)=\prod_{j=1}^n(ta_j+(1-t)b_j).$$Since $|a_j|\le 1$ and $|b_j|\le 1$ the product rule shows that $$|f'(t)|\le\sum|a_j-b_j|\quad(0\le t\le 1),$$hence $$|f(1)-f(0)|\le\sum|a_j-b_j|.$$