$c_0$ is isometrically isomorphic to a subspace of $C[0,1]$
My attempt (with aid of a hint): Define $T: c_0 \to C[0,1]$ as $(x_n)\mapsto f$ where $f$ is the linear extension of the function $\frac{1}{n}\mapsto x_n$.
Well-defined. Since $f$ is piecewise-continuous, so by pasting lemma, $f$ is continuous and hence $f\in C[0,1]$.
Bounded. $\|T(x_n)\|=\|f\|\le \sup_n |x_n| = \|(x_n)\|$. Thus $\|T\|\le 1$.
Also, $\|f(e_1)\|=\sup_{x\in[1/2,1]}2x-1=1$, so $ \|T\|\ge 1$.
My doubts: How to show that $T$ is linear? I'm also unable to show that $T$ is an isometry. Lastly, to show boundedness, it is geometrically obvious to me that $\|f\|\le \sup_n |x_n|$ but is there a way to rigorously show this?
Linearity: Let $x^1, x^2 \in c_0$. We have, for $y \in (0,1]$ there exists $n$ such that $y \in [1/(n+1),1/n]$ and \begin{align} T(x^1 + x^2)(y) &= f(y)\\ &= \frac{(x^1 + x^2)_{n+1} - (x^1 + x^2)_{n}}{\frac{1}{n+1} - \frac{1}{n}} (y - 1/n) + (x^1 + x^2)_n \\ &= \left(\frac{x^1_{n+1} - x^1_{n}}{\frac{1}{n+1} - \frac{1}{n}} (y - 1/n) + x^1_n \right) + \left(\frac{x^2_{n+1} - x^2_{n}}{\frac{1}{n+1} - \frac{1}{n}} (y - 1/n) + x^2_n \right)\\ &= f^1(y) + f^2(y) \end{align} which prove the linearity (note that if $y = 0$ then $f(y) = 0 = f^1(y) + f^2(y)$).
Isometry: For this part, just remark that, if $f \neq 0$, $$\|f\| = \sup_{y \in [0,1]}|f(y)| = \sup_{n \in \mathbb N_0} \sup_{y \in [1/(n+1),1/n]}|f(y)|.$$ As $$\sup_{y \in [1/(n+1),1/n]}|f(y)|$$ is either $x_n$ or $x_{n+1}$, you can conclude.