Some problem of function matric space I have some problem doing my homework

Let $|\cdot |$ be define as $|f| = max\{|f(t)|:t\in [0,1]\}$
Define an integral transformation $T:C^0\to C^0$ by$$T(f)(x)=\int_0^xf(t)\,dt$$
(a) Show that $T$ is continuous and find it's norm.
(b)Let $f_n(t) = cos(nt), n=1,2,3\ldots$, What is $T(f_n)$?
(c)Is the set of function $K = \{T(f_n):n\in \Bbb N\}$closed?bounded?compact?
(d)Is $T(K)$ compact? How about it's closure?

I'm not sure if I get the right idea of what the exercise is talking about, so I write the whole text down.
My works:
(a) It's continuous because $T$ maps continuous $f$ only, so I can say the integral of $f$ is also continuous.
About it's norm, I'm not sure which one I pick up the maximum, $f$ or $x$. I suppose is $x$ because I don't know how to pick $f\in [0,1]$. So my answer is $$|T(f)(x)| = max\{|T(f)(x)|:t\in [0,1]\} = max\{|\int_0^tf(s)\,ds|:t\in [0,1]\}$$
(b)I think it's $\frac1n\sin nx$
(c)not close because I take the seq$=\{\frac1 2 \sin 2x,\frac 1 3 \sin 3x,\ldots\}$, the seq goes to $0$ but $0$ is not in $K$
bounded yes, all the norm in $K$ is less than $1$, so bounded by 1.
compact, I don't know but I think it's right, because $K$ is like $\{\frac1n:n\in\Bbb N\}$, all guessing, no big deal...
(d)$T(K)=\{\frac1n-\frac1n\cos nx : n\in \Bbb N\}$... and then?
It's possible that what I did is totally wrong, truly possible, so thank you
 A: (a) $T$ being continuous has nothing to do with $T$ mapping continuous functions. $T\colon C^0 \to C^0$ is continuous iff it is bounded, which means that the number $\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$
$$ \norm T := \sup_{\abs f \le 1} \abs{Tf} $$
(the norm of $T$) is finite. Let $f \in C^0$ with $\abs f \le 1$. Then we have
\begin{align*}
  \abs{Tf} &= \sup_{x \in [0,1]} \abs{Tf(x)}\\
           &= \sup_{x\in [0,1]} \abs{\int_0^x f(t)\, dt}\\
           &\le \sup_{x\in [0,1]} \int_0^x \abs{f(t)}\, dt\\
           &\le \sup_{x\in [0,1]}\int_0^x \abs f\, dt\\
           &= \sup_{x\in[0,1]} x \abs f\\
           &= \abs f \le 1
\end{align*}
Hence $\norm T \le 1$ and $T$ is continuous with $\norm T \le 1$. To show that the norm of $T$ equals $1$, we can give a function $f \in C^0$, with $\abs f = \abs {Tf} = 1$, for example $f = 1$, then $T1 = {\rm id}$, and $\abs 1 = \abs{\rm id} = 1$, so
$$ \norm T = \sup_{\abs f \le 1}\abs{Tf} \ge \abs{T1} = \abs{\rm id} = 1 $$
Hence $\norm T = 1$.
(b) You are right, $Tf_n(x) = \frac 1n \sin nx$, each $n \in \mathbf N$, $x \in [0,1]$.
(c) Right, the sequence $(Tf_n) \in K^{\mathbf N}$ converges to $0 \in C^0$, as 
$$ \abs{Tf_n} = \frac 1n \sup_{x \in [0,1]}\abs{\sin(nx)} = \frac 1n \to 0 $$
but $0 \not\in K$, but $K$ is bounded, as 
$$ \sup_{f \in K} \abs{f} = \sup_{n\in \mathbf N}\abs{Tf_n} = \sup_{n\in \mathbf N}\frac 1n = 1 $$
As compact sets are closed, $K$ being not closed, cannot be compact.
(d) We have for $n\in \mathbf N$, $x \in [0,1]$, that 
$$ T^2f_n(x) = T(Tf_n)(x) = \int_0^x \frac 1n \sin(nt)\,dt = \frac 1{n^2}(1 - \cos nx) $$
$T(K)$ is again not closed (note that $T^2f_n \to 0$, as $Tf_n \to 0$ and $T$ is continuous), hence not compact. The closure of $TK$ consits of a convergent sequence (namely $(T^2f_n)$) and its limit, and is hence compact (as such sets are compact in every metric space: Given an open cover, one of the sets must contain the limit [here: 0], and by convergence hence all but finitely of the $T^2f_n$, giving a finite subcover).
