Consider the real vector space of all functions from $\mathbb{R} \rightarrow \mathbb{R}$. Suppose we have the subset defined by: $$f(0)=1+f(1)$$
Now I want to prove that this is not a subspace. My first observation would be that $$(f+ g)(0)=1+(f+ g)(1)$$, However, when we add that two functions separately we get: $$f(0) +g(0)=1+f(1) +1 +g(1) =2 + (f + g) (1)$$ Is this reasoning valid? It simply means that we cannot add elements and stay in the same set.
I also notice that the defined addition is commutative, associative, but there also would be no identity function satisfying this property, as the zero function is also not in there take $f(x)=0$ then $ f(0)=0 \not = 1+0$. Are there other properties that fail, which I am missing?
You proved more than you needed to prove. Yes, if you sum any two elements of the set, the sum doesn't belong to it. A concrete example will do then. For instance, take $f(x)=g(x)=-x+1$. Then $f$ and $g$ belong to the set, but not their sum.
Or you can just note that the null function (which is the $0$ of the whole space) does not belong to the set.