This is an exercise in the "lecture notes in functional analysis" by bressan. asking to show whether the following operators are bounded and compact on the Banach space $X = C([0, 1])$, (I try to write all the detail steps)
$\textbf{- $\Lambda f(x)=f(sinx)$ }$
bdd : $|\Lambda f(x)|=|f(sinx)| \le \|f\| \implies \|\Lambda f\| = \|f\| \implies \|\Lambda\| = \sup_{f}\frac{\|\Lambda f\|}{\|f\|} = \frac{{\|f\|}}{{\|f\|}} =1 $
not compact : let $f_n(x) =x^n$ , then $\Lambda f_n(x)= [sin(x)]^n$ , if $sin(x)=1 , \Lambda f_n(x) \to 1$ , otherwise $\ \ \Lambda f_n(x) \to 0 $. So there would be no converging subsequence for this example.
$\textbf{-$\Lambda f(x)=xf(x)$}$
bdd : $|\Lambda f(x)|=|xf(x)| \le \|x\|\|f\| \implies \|\Lambda f\| = \sup_{x}\frac{\|x\|\|f\| }{\|x\|}=\|f\| \implies \|\Lambda\| = \sup_{f}\frac{\|\Lambda f\|}{\|f\|} = \frac{{\|f\|}}{{\|f\|}} =1 $
not compact : let $f_n(x) =x^n$ , then $\Lambda f_n(x)= x^{n+1}$ , if $x=1 , \Lambda f_n(x) \to 1$ , otherwise $\ \ \Lambda f_n(x) \to 0 $. So there would be no converging subsequence for this example.
$\textbf{-$\Lambda f(x)=xf(0)+\int_0^1 f(s)ds$}$
bdd : $|\Lambda f(x)|=|xf(0)+\int_0^1 f(s)ds|\le 2\|f\| \implies \|\Lambda f\| = 2\|f\| \implies \|\Lambda\| = \sup_{f}\frac{\|\Lambda f\|}{\|f\|} = \frac{2\|f\|}{\|f\|} =2 $
compact : we always have $\Lambda f(x)=xf(0)+\int_0^1 f(s)ds = P(\text{at most degree 1})+\text{constant}$ , so the range of $\Lambda f(x)$ is at most two. so it is a compact operator.
$\textbf{-$\Lambda f(x)= y(x)$}$ , where y(·) is the solution to the Cauchy problem $y'(x)+y(x)=f(x) , \ \ y(0)=0$
bdd : we have that by above $\Lambda f(x) = \int_0^x e^{y-x}f(y)dy \implies |\Lambda f(x)| = |\int_0^x e^{y-x}f(y)dy| \le \|f\| \implies \|\Lambda\| = 1 $
compact :by the compactness of an integral operator.
For the first one, $\Lambda f(x)=f(\sin x)$. Note that when $x\in[0,1]$, we have $\sin x\in[0,\sin1]\subset[0,1]$. This is because $1<\pi/2$ so $\sin 1<1$. As a result, for any $f$, it follows that $\sup_{x\in[0,\sin 1]}|f(x)|\leq \sup_{x\in[0,1]}|f(x)|=||f||.$
Bounded: I agree that $|\Lambda f(x)|\leq ||f||$ for all $x\in [0,1]$ so $||\Lambda f||\leq ||f|| \implies ||\Lambda||\leq 1$. This, of course, finishes the proof. But in your answer, you claim that $||\Lambda||=1$. You really should show that there exists $f$ such that $||\Lambda f||=||f||$. This is easy, as just choose $f=1_{[0,1]}$.
(not) Compact: Your counterexample doesn't prove it. This is because for $f_n(x)=x^n$, we have $\Lambda f_n(x)=(\sin x)^n$. Therefore, $||\Lambda f_n||= (\sin 1)^n \to 0$ since $\sin 1<1$. Hence, obviously, a convergent subsequence exists. There is an easy way to modify your counterexample to make your main idea hold true.
For the second one, $\Lambda f(x)=xf(x)$, you can say $|\Lambda f(x)|\leq |x|||f|| \leq ||f||$ so $||\Lambda|| \leq 1$. For the other inequality, choose $f(x)=1$ again. This time I agree with your proof of (not) compactness.
In general, you keep making similar mistakes. I suggest you read more about how to prove the norm of an operator equals a number.