For convenience, define $f(s,t) = \prod_{t=i}^s \left(1 + \frac{1}{4} x_i\right)$ and $g(s,t) = \prod_{t=i}^s \left(1 + \frac{1}{2} x_i\right)$.
Note that since $x_i \leq 0$, both $f(s,t)$ and $g(s,t)$ are monotonically decreasing in $s$ and monotonically increasing in $t$. Therefore,
$$
f(S,1) = \inf_{s \in [1,S]} \inf_{t \in [1,s]} f(s,t) \quad \text{and} \quad g(S,1) = \inf_{s \in [1,S]} \inf_{t \in [1,s]} g(s,t)\enspace.
$$
Since $x_j \leq 0$ for all $j$, we have
$$
\prod_{i \in A} \left( 1 + \frac{1}{4} x_i \right) \geq \prod_{i \in A} \left( 1 + \frac{1}{2} x_i \right) \enspace,
$$
for any set $A \subset ( \mathbb{N} \cap [1,S] )$. To get a strict inequality, pick any $x_k$ such that $x_k < 0$ (and $k \in [1,S]$) and define $A = ((\mathbb{N} \cap [1,S]) \setminus \{x_k\})$. Clearly,
\begin{align*}
f(S,1) = \prod_{i=1}^S \left( 1 + \frac{1}{4} x_i \right)
& = \left( 1 + \frac{1}{4} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{4} x_i \right) \\
& \geq
\left( 1 + \frac{1}{4} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{2} x_i \right) \\
& > \left( 1 + \frac{1}{2} x_k \right) \prod_{i \neq k}^S \left( 1 + \frac{1}{2} x_i \right)
= \prod_{i=1}^S \left( 1 + \frac{1}{4} x_i \right) = g(S,1)
\enspace.
\end{align*}
And we are done.