Answer
See below
Work Step by Step
Given $V$ is the vector space of all real-valued functions defined on the interval $[-\infty, \infty]$, and $S$ is the subset of $V$
We can write set $S$ as $S=\{f \in V:f(-x)=f(x),\forall x \in (-\infty, \infty)\}$.
We can notice that $f:(-\infty, \infty) \rightarrow R$ given by $f(x)=0 \forall x \in [a-\infty, \infty] \in S$. Hence, $S$ is nonempty (1)
Let $f,g \in S$. Then we have $f(-x)=f(x),g(-x)=g(x)$
Obtain $(f+g)(-x)\\=f(-x)+g(-x)\\=f(x)+g(x)\\=(f+g)(x) \forall x \in (-\infty, \infty)$
Since $(f+g)(x) \in S$, $f+g$ is closed under addition (2)
Let $f \in S$ and $c$ be a scalar. Then we have $f(-x)=f(x) \forall x\in (-\infty, \infty)$
Obtain $(cf)(-x)\\=cf(-x)\\=cf(x)\\=(cf)(x) \forall x \in (-\infty, \infty)$
Since $(cf)(x) \in S$, $cf$ is closed under scalar multiplication (3)
From (1), (2), (3), $S$ is a subspace of $V$
