I know that the above statement is true because:
Proof: Since $\ln(1-p_m) \leq -p_m$,
$$\prod_{m=1}^n(1-p_m) = \prod_{m=1}^n e^{\ln(1-p_m)} \leq \prod_{m=1}^n e^{-p_{m}} = e^{-\sum_{m=1}^n p_m}$$
If we take the limit on both sides, we get
$$\lim_{n\rightarrow \infty} \prod_{m=1}^n (1-p_m) \leq \lim_{n\rightarrow \infty} e^{-\sum_{m=1}^n p_m} = 0$$
But how can I prove this statement with the use of the BC lemmas? can someone help me construct this proof please because I know this statement is true and can be used to prove the BC Lemma but how can I use the BC lemma to show this?
First of all, it appears that you have proven only one direction of the statement (the if statement). Let us prove both directions at once with the use of the Borel Cantelli lemma. Recall that one of the statements of the lemma is that for a sequence of independent events $E_n$:
$$\sum_{n \geq 1} \mathbb{P}(E_n) = + \infty \qquad \iff \qquad \mathbb{P}(\limsup_{n \to \infty} E_n) = 1.$$
We can then choose a sequence of independent events $E_n$ such that $\mathbb{P}(E_n)= p_n$. Note moreover that:
$$\mathbb{P}(\limsup_{n \to \infty} E_n) = \lim_{n \to \infty} \mathbb{P}(\cup_{m \geq n} E_m) = \lim_{n \to \infty} (1 - \mathbb{P}(\cap_{m \geq n} E_m^c)) = 1 - \lim_{n \to \infty} \prod_{m \geq n }(1 - p_m)$$
We then find from above that
$$ \mathbb{P}(\limsup_{n \to \infty} E_n) = 1 \qquad \iff \qquad \lim_{n \to \infty} \prod_{m \geq n }(1 - p_m) = 0 \qquad \iff \qquad \prod_{n \geq 1} (1 - p_n) = 0. $$
This is the required result.