= Solution
First recall from the <least-upper-bound property> that every <real interval> is connected. For completeness, if $[a,b]=U\cup V$ were a separation with $a\in U$, let $s$ be the <supremum> of the points $t$ for which $[a,t]$ remains in $U$. Relative openness rules out both $s\in U$ and $s\in V$: in the first case one can move slightly to the right while staying in $U$, and in the second one finds points of $V$ slightly to the left of $s$, contrary to the definition of $s$. Thus no separation exists.
Part (a) now makes $g([0,1])$ a <connected subset of the real line>. Such a set is an interval: if it contained $u<v$ but omitted $y\in(u,v)$, its intersections with $(-\infty,y)$ and $(y,\infty)$ would separate it. Since $g([0,1])$ contains $g(0)$ and $g(1)$, it consequently contains every number between them. Therefore for each such $y$ there is an $x\in[0,1]$ with $\boxed{g(x)=y}$. This derives the <intermediate value theorem> from connectedness.
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