Compact subset of space of Continuous functions

Let $f_n(t)=(n+1)t^n$. Then $f_n \in K$ for all $n$. Can you proceed ?

Consider the sequence $(f_n)_n$ given by $f_n(x) = 1+n\sin(2\pi x)$. We have $(f_n)_n \subseteq K$ but $$\|f_n\|_\infty \ge f_n\left(\frac14\right) = 1+n\sin\left(\frac\pi2\right) = 1+n $$

Hence $K$ isn't bounded so it cannot be compact.

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An alternative argument: define a linear functional $\phi : C[0,1] \to \mathbb{R}$ as $\phi(f) = \int_0^{1/2}f(t)\,dt$. We have that $\phi$ is bounded and hence continuous with respect to the supremum norm.

If $K$ were compact, $\phi|_K$ would be a bounded function. However, for the functions $(f_n)_n$ above we have $$\phi(f_n) = \int_0^{1/2}f_n(t)\,dt = \frac12 + \frac{n}\pi$$ which is a contradiction.