= Solution
Let $u,v\in f(I)$ with $u<y<v$. Choose preimages $x_u,x_v\in I$, and let $a=\min(x_u,x_v)$, $b=\max(x_u,x_v)$. The defining property of a <real interval> ensures $[a,b]\subseteq I$. The restriction of $f$ to this <closed interval> is <continuous>, and its endpoint values are $u,v$ in one order or the other. The <intermediate value theorem> supplies $c\in[a,b]\subseteq I$ with $f(c)=y$. Thus every value between two image values lies in the image, proving
$$
\boxed{f(I)\text{ is a real interval}.}
$$
The <continuous image of a real interval> result does not require $I$ itself to be closed or bounded; only the segment between the chosen preimages is used. Empty and single-point images also satisfy the interval property.
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