It is the space of all continuous on $[0,1]$ functions with the uniform metric. We need to show that the space is separable (and, therefore, being metric, it is also second-countable). Consider any finite subset of $[0,1]\cap\mathbb{Q}$ that includes both end points. There are countably many such sets. Now consider all functions that takes rational values at the points of such a finite set and are linear between the points. They are all continuous functions (use the Pasting Lemma from §18 or the fact that the function is bounded, therefore, the image lies within a compact Hausdorff space and the result from §26 tells us that it is continuous iff its graph is closed) well-defined uniquely by the condition above (all points in a finite set are isolated and for each point not in the set there is the closest point below and another above), and there are countably many of such functions. All we need to show that any continuous function can be approximated by functions from this collection. This is quite easy to show using the results from §27. A continuous function from a compact metric space to a metric space is uniformly continuous. For a given $\epsilon$ find $\delta$ such that if $|x-x'|<\delta$ then $|f(x)-f(x')|<\epsilon/4$ . Take a finite set of points $\{x_n\}$ of $[0,1]$ such that the distance between a pair of successive point is less than $\delta$ . For each point $x_n$ find a rational number $|q_n-f(x_n)|<\epsilon/6$ and construct the dense function $q$ . Note that $|q_{n+1}-q_n|<7\epsilon/12$ . Then for any $x\in[q_n,q_{n+1}]$ we have $|f(x)-q(x)|<|f(x)-f(x_n)|+|f(x_n)-q_n|+|q(x)-q_n|<$ $\epsilon/4+\epsilon/6+|q_{n+1}-q_n|<\epsilon$ .