Prove that $\{f : f$ is a constant on the interval $[0,1]\}$ is a subspace of $\mathscr{F}(R)$.
This doesn't seem true. Let $U=\{f : f$ is a constant on the interval $[0,1]\}$. Obviously $U\subseteq \mathscr{F}(R)$. For every $k\in R$ and $f\in U$, $kf(x)$ is not necessarily a constant on the interval $[0,1]$. Since $U$ is not closed with respective to scalar multiplication, $U$ is not a subspace of $\mathscr{F}(R)$.
What am I missing?
I guess $R=\Bbb R$ and $V:=\mathcal F(R)$ denotes the $\Bbb R$-vector space of functions $\Bbb R \rightarrow \Bbb R$ with pointwise scalar multiplication and addition.
Then $U:=\{f:\Bbb R \rightarrow \Bbb R\mid f \text{ constant on }[0,1]\}$ is a subspace of $V$. Indeed, given $f,g\in U$ (with $f(x)=k, g(x)=l$ for all $x\in[0,1]$) and $\alpha \in \Bbb R$ we find that $$(\alpha f)(x) = \alpha f(x) = \alpha k$$ and $$(f+g)(x) = f(x)+g(x) = k+l$$ for all $x\in[0,1]$. Since $0$ is constant on $[0,1]$ this shows the claim.