Proof. First note that if $0 \not\in A$ or $1 \not\in A$, $C(A; E)$ embeds isometrically into $C(A \cup \bracs{0, 1}; E)$ through extension by $0$. Thus assume without loss of generality that $A$ contains the endpoints $0$ and $1$.

Let $U = [0, 1] \setminus A$, then there exists $\seq{(a_n, b_n)}\subset [0, 1]^{2}$ such that $U = \bigsqcup_{n \in \natp}(a_{n}, b_{n})$. For each $f \in C(A; E)$, let

then the mapping $f \mapsto Tf$ is an isometric embedding into $E^{[0, 1]}$ with respect to the uniform norm.

Since $U$ is open and $Tf$ is affine on each component of $U$, $Tf$ is continuous on $U$. It remains to show that $Tf$ is continuous on $A$. Let $x \in A$ and $\eps > 0$, then there exists $\delta > 0$ such that $\norm{f(y) - f(x)}_{E} < \eps$ for all $y \in (x -\delta, x + \delta) \cap A$. Now, a case analysis:

1. (1)

   If there exists $y \in (x - \delta, x) \cap A$, then for any $z \in U \cap (y, x)$, there exists $n \in \natp$ such that $(a_{n}, b_{n}) \subset (y, x)$ and $z \in (a_{n}, b_{n})$. In which case, since $\norm{f(a_n) - f(x)}_{E} < \eps$ and $\norm{f(b_n) - f(x)}_{E} < \eps$, $\norm{Tf(z) - Tf(x)}_{E} < \eps$. Thus $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (y, x)$.

2. (2)

   Otherwise, $x = 0$ or $Tf|_{(x - \delta, x)}$ is an affine function. Either way, there exists $y \in (x - \delta, x)$ such that $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (y, x) \cap [0, 1]$.

Thus there exists $y \in (x - \delta, x)$ with $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (y, x) \cap [0, 1]$. Similarly, there exists $y' \in (x, x + \delta)$ with $\norm{Tf(z) - Tf(x)}_{E} < \eps$ for all $z \in (x, y') \cap [0, 1]$. Therefore $Tf$ is continuous at $x$. Since this holds for all $x \in U$ and $x \in A$, $Tf \in C([0, 1]; E)$.$\square$