Answer
See below
Work Step by Step
Given $S$ is the subset of $V$
$\rightarrow S=f,f:[a,b]\rightarrow a,f(a)=5f(b)$ such as $f=0 \in S\\
\rightarrow f \ne g$
then $(f+g)(a)\\=f(a)+g(a)\\=5f(b)+5g(b)\\=5(f(b)+g(b))\\=5((f+g)(b))$
Since $f+g \in S$, $f+g$ is closed under addition.
Let $k$ be a scalar
Obtain $(kf)(a)=kf(a)=k5f(b)=5(kf)(b)$
We can see that $kf \in S$. Hence $kf$ is closed under scalar multiplication (2)
From (1)(2), $S$ is a subspace of $V$
