Showing that a shape $[a_1,b_1]\times ... \times [a_n,b_n]$ in $R^n$ is convex $\times$ being the topological product (ie, the book describes the shape as "rectangles in $R^n$, as the combination of 2 lines makes a rectangle, and adding a third makes a prism, and so on in higher dimensions), and convex being that for any $a$ and $b$ in the shape, all elements in the line $[a,b]$ are contained in the shape.
I tried essentially stating that, firstly, shapes as defined in the title in $R$ are trivially convex. Then, if we say that such a shape $[a_1,b_1]\times ... \times [a_{n-1},b_{n-1}]$ in $R^{n-1}$ is convex, then for every line $[x,y]$ in $[a_1,b_1]\times ... \times [a_n,b_n]$, it can be written as $[(x_1,...,x_{n-1}),(y_1,...,y_{n-1})]\times[x_n,y_n]$ which are lines fully contained in $[a_1,b_1]\times ... \times [a_{n-1},b_{n-1}]$ and $[a_n, b_n]$, so [x,y] is fully contained in $[a_1,b_1]\times ... \times [a_n,b_n]$.
However, this proof seems incredibly weak and circular at best to me, but I always have trouble knowing where to start with hard proofs for something like this which seems so conceptually intuitive.
 A: The shape you are describing here is a "higher dimensional cuboid", you should use this as intuition for your proof, the result is clearly true in the 1 and two dimensional cases, but you must generalise this proof to higher dimension.
Recall that a set $X\subset\Bbb R^n$ is convex if for each $\mathbf x,\mathbf y\in X$ and each $t\in[0,1]$, we have $t\mathbf x+(1-t)\mathbf y\in X$.
Assume that $a_j<b_j$ for all $j$, to avoid degenerate cases (these will also need to be dealt with). You need to check that line segments connecting any two elements of the set are again contained in the set.
To do this, let $\mathbf x,\mathbf y\in[a_1,b_1]\times\dots\times[a_n,b_n]$. We must show that, for each $t\in[0,1]$, we have $t\mathbf x+(1-t)\mathbf y\in[a_1,b_1]\times\dots\times[a_n,b_n]$. Indeed, for each $1\leq j\leq n$ and $t\in[0,1]$, we have $tx_j+(1-t)y_j\in[a_j,b_j]$ (justify!) where $x_j$ is the $j$-th component of $\mathbf x$ and $y_j$ is the $j$-th component of $\mathbf y$. From this it follows that $t\mathbf x+(1-t)\mathbf y\in[a_1,b_1]\times\dots\times[a_n,b_n]$ (this requires a little argument too).
