You can prove $V$ is a subspace of $F$ the same way you prove any subspace is a subspace: show it is closed under addition and under scaling. Explicitly, if $f,g$ are in $V$, show $f+g$ is in $V$, and $\alpha f$ is in $V$. If you don't know the defintions of addition or scaling of functions, you can ask, but I'm betting you've seen it.
The second half of the question is garbled. You wrote "f(1)" which as written is in $\mathbb{R}$, so it's not in $F$ at all. If you meant $"f1"$ instead, then the statement is trivial, because $x$ is surely in the span of $\{x,y,z,\dots\}$... can you see how to arrange coefficients on the three functions so that the result is f1?
Note The OP improved the typesetting by a lot since this was posted, but fortunately the answers still apply.
I'll add that as long as you show $T$ is linear, it isn't important how $T$ is defined: $\{x\in F\mid Tx=0\}$ is always going to be a subspace (because it's the kernel of a linear operator.)