Show that the following operator is not a surjection. 
"Take $X=C([0,1])$ with the uniform norm, $\|f\|=\sup_{x\in[0,1]}|f(x)|$, and define the operator $T:X\to X$ by,
  $$T(f)(x)=f(x)-\int_0^1f(s)ds$$
  Show that $T$ is not a surjection".

Here is what I have done:
Saying that $T$ is not a surjection means that $\exists\, g\in X:\nexists f\in X:Tf=g$
So we consider,
$$T(f)(x)=f(x)-\int_0^1f(s)ds=g(x)$$
$$\implies\int_0^1f(x)dx-\int_0^1\int_0^1f(s)ds\,dx=\int_0^1g(x)dx$$
$$\implies0=\int_0^1g(x)dx$$
In the solutions I have, it says that two functions are equal if we obtain the same value after evaluating their integrals over the interval. Is this enough to show that $T$ is not a surjection? I am not entirely convinced.
Clearly the only $g\in X$ for which this holds is $g=0$, the zero function in $X$. But since $g\in X$ was arbitrary, and not every $g\in C([0,1])$ satisfies $\int_0^1g(x)dx=0$, $T$ is not a surjection?
Is my thinking here correct?
 A: 
In the solutions I have, it says that two functions are equal if we obtain the same value after evaluating their integrals over the interval. Is this enough to show that $T$ is not a surjection? I am not entirely convinced.

That statement is backwards: If two functions are equal, then obviously they have the same integral, but there are plenty of non-zero functions in $C([0,1])$ whose integral is $0$. For example, $f(x) = \sin(2\pi x)$. Having the same integral is nowhere near enough information to say that two functions are equal.

Clearly the only $g\in X$ for which this holds is $g=0$, the zero function in $X$. 

Clearly not.

But since $g\in X$ was arbitrary, and not every $g\in C([0,1])$ satisfies $\int_0^1g(x)dx=0$, $T$ is not a surjection?

$g$ is not arbitrary. To show that $T$ is not a surjection, you just have to find a single function that is not in its image. You don't have to prove it for a class of functions, but just show that there is at least one function for which this condition fails.
As Joey Zou has pointed out, you have shown that for $g = Tf$, it must be true that $\int_0^1 g(x)dx = 0$. So what you need to do now is give a specific example of a function in $C([0,1])$ whose integral is not $0$. Then that function cannot be the image of any $f$, so $T$ cannot be surjective.
