Question: Is the subset $ B$ of vector space $ F(\mathbb{R},\mathbb{R})$ the subspace of this vector space?;
$ B=\{f: f(7)=2+f(1)\}$ 

My prove:

let $g(x),h(x) \subset B$

1) $(\alpha\*g)(7)=\alpha \* g(7) =\alpha \*(2+g(1))=2\*\alpha+ (\alpha\*g)(1) \not\subset B $ 

2) $(g+h)(7)=g(7)+h(7)=4 + g(2) +h(2) = 4 + (h+g)(2) \not\subset B $ 

To show that it's not a subspace it's enough to prove that one of conditions 1),2) is false and both are false, so $B$ is not a subspace of $ F(\mathbb{R},\mathbb{R})$.

Is it correct?

Answer: 
That is the same error as in $1=\sqrt{1}=\sqrt{(-1)^2}=-1$ 

That is based on the fact: $1^\frac{1}{2} = \sqrt{1} = \pm 1$  

The same is true for $1^\frac{1}{6} = \sqrt[6]{1} = \pm 1$  

So you get $-1=\pm 1$ and that is true.
The mistake lies in assuming that $(1)^{\frac{1}{6}}=1$, this is because, $1$ has two sixth roots in the field of real numbers, $1,-1$