Is the Banach space of continuous functions $I \to \ell^2$ separable? Inspired by this question.
Consider the vector space $V$ of all continuous functions $I \to \ell^2$ for $I=[0,1]$ the closed unit interval and $\ell^2$ the Hilbert space of all square-summable sequences. Under the $\sup$ norm $V$ becomes a Banach space.
I would like to know is $V$ separable.
Suppose $p_1,p_2,\ldots : I \to \ell^2$ form a countable dense subset. Believing the answer is no, we might try a diagonal argument to construct some $f$ with all $\|f-p_n\|>1$. One idea is to demand for $x = 1-1/n$ that the $n$-th coordinate of $f$ is equal to $p_n(x)+1$. Then we have a lot of freedom choosing the rest of the values for $f$. The problem with the approach is we have to choose those values so the limit $f(1-1/n) \to f(1)$ and that's hard to do knowing nothing about  $p_1,p_2,\ldots $.
 A: Yes, $V$ is separable.  Let $Q$ be a countable dense subset of $\ell^2$ and let $D$ be the set of functions $f:I\to \ell^2$ such that there exists $n\in\mathbb{Z}_+$ such that $f(k/n)\in Q$ for $k=0,1,\dots,n$ and $f$ interpolates linearly between these values.  Then $D$ is countable.
To prove that $D$ is dense, let $f\in V$ and $\epsilon>0$.  Since $I$ is compact, $f$ is uniformly continuous, so we can pick $n$ such that $|s-t|\leq 1/n$ implies $\|f(s)-f(t)\|<\epsilon$ for all $s,t\in I$.  Now let $g$ be such that $g(k/n)\in Q$ and $\|g(k/n)-f(k/n)\|<\epsilon$ for $k=0,1,\dots,n$ and $g$ interpolates linearly in between.  Let $t\in I$, and let $k$ be such that $t\in [k/n,(k+1)/n]$.  Note that $f(k/n)$ is within $\epsilon$ of $f((k+1)/n)$ by our choice of $n$, and so $g(k/n)$ is within $3\epsilon$ of $g((k+1)/n)$.  Since $g$ is linear from $k/n$ to $(k+1)/n$, it follows that $g(t)$ is also within $3\epsilon$ of $g(k/n)$, and hence within $4\epsilon$ of $f(k/n)$.  Finally, $f(k/n)$ is within $\epsilon$ of $f(t)$ again by our choice of $n$, so $g(t)$ is within $5\epsilon$ of $f(t)$.
Since $t\in I$ was arbitrary, this shows that $\|f-g\|\leq 5\epsilon$ in $V$.  Since $g\in D$ and $\epsilon>0$ was arbitrary, this shows that $D$ is dense in $V$.
