Let $T$ can take on values $t_1,t_2,\ldots,$ with $0\le t_1\le t_2,\ldots,$ and let the probability function be
$$f(t_j)=Pr(T=t_j),\quad j=1,2,\ldots$$
The survivor function is then $$S(t)=Pr(T\ge t)=\sum_{j:t_j\ge t}f(t_j)$$
The discrete time hazard function is defined as : $$h(t_j)=\frac{f(t_j)}{S(t_j)}=1-\frac{S(t_j)}{S(t_{j+1})},\quad j=1,2,\ldots$$
since $f(t_j)=S(t_j)-S(t_{j+1})$.
Using the discrete nature of $T$, here is a heuristic way to derive the survival function at time $t_j$:
$$ \begin{align*}S(t_j) & = \text{surviving past time }t_j \\ & = \text{surviving past time } t_{j-1} \text{ and not dying at } t_j \\ & = \Pr\big(\text{not dying at }t_j \,\big|\, \text{surviving past } t_{j-1}\big) \times \Pr\big(\text{surviving past time } t_{j-1}\big) \\ & = \Pr\big(\text{not dying at }t_j \,\big|\, \text{survival up to } t_j\big) \times \Pr\big(\text{surviving past time } t_{j-1}\big) \\ & = [1 - h(t_j)] \times S(t_{j - 1}) \end{align*}$$
For all $j=1,2,\ldots$ we have $$ 1-h(t_j)=\frac{P(T\geq t_j)-P(T=t_j)}{P(T\geq t_j)}=\frac{P(T\geq t_{j+1})}{P(T\geq t_j)}. $$ Now, if $t>0$, then we choose the largest $t_k$ such that $t_k\leq t$. Thus $$ \prod_{j:t_j\leq t}[1-h(t_j)]=\prod_{j:t_j\leq t_k}[1-h_j(t)]=\prod_{j=1}^k\frac{P(T\geq t_{j+1})}{P(T\geq t_j)}. $$ Noting that this is a telescoping product, we end up with $$ \prod_{j:t_j\leq t}[1-h(t_j)]=\frac{P(T\geq t_{k+1})}{P(T\geq t_1)}=P(T>t_k)=S(t_k)=S(t). $$
