Answer
See below
Work Step by Step
Given: $V=C[a,b]\\
S=\{f \in V:f(a)=2f(b)\}$
Take $f \in C[a,b]$ such that $f(x)=0 \forall x \in [a,b]$.
If we take $f(a)=0=2.0=2f(b) \rightarrow f \in W$
We can see $W$ is nonempty (1)
Let $f(a)=2f(b)\\
g(a)=2g(b)$
then $(f+g)(a)=f(a)+g(a)=2f(b)+2g(b)=2(f(b)+g(b))=2(f+g)(b)\\
\rightarrow f+g \in C[a,b]\\
\rightarrow f+g \in W$
Hence, $f+g1$ is closed under addition multiplication (2)
Let $k$ be a scalar
Obtain $(kf)(a)=kf(a)=k(2f(b))=2kf(b)=2(kf)(b)\\
\rightarrow kf \in C[a,b]\\
\rightarrow kf \in W$
Hence, $kf$ is closed under scalar multiplication (3)
From (1)(2)(3), $W$ is a subspace of $C[a,b]$
