Answer
See below
Work Step by Step
Given $V=C^2(I)$
We can write set $S$ as $S=\{p \in C^2(I):y''+2y'-y=0\}$.
Assume that $y_0 \in C^2(I)$ such as $y_0(t) =0 \forall t \in I$. By then we have $y''_0(t)=0,y_0'(t)=0 \forall t \in I$
We can see that $y_0''(t)+2y_0'(t)-y_0(t)=0 \forall t\in I$. Hence, $S$ is nonempty (1)
Let $y_1,y_2 \in S$ then $y_1''+2y_1'-y_1=0,y_2''+2y_2'-y_2 =0$
Obtain $(y_1+y_2)''(t)+(y_1+y_2)'(t)-(y_1+y_2)(t)\\=y_1''(t)+y_2''(t)+y_1'(t)+y_2'(t)-y_1(t)-y_2(t)\\=[y_1''(t)+y_1'(t)-y_1(t)]+[y_2''(t)+y_2'(t)-y_2(t)]\\=0+0=0 \forall t \in I$
Since $y_1+y_2 \in S$, $y_1+y_2$ is closed under addition (2)
Let $y \in S$ and $c$ be a scalar
Obtain $(cy'')(t)+2(cy)'(t)-(cy)(t)\\=cy''(t)+2cy'(t)-cy(t)\\=c[y''(t)+2y'(t)-y(t)]\\=c(0)=0 \forall t \in I$
Since $cy \in S$, $cy$ is closed under scalar multiplication (3)
From (1), (2), (3), $S$ is a subspace of $C^2(I)$
