$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Let $Y$ denote the interior of $X$. Then indeed $g:=\na f$ is a homeomorphism of $Y$ onto $g(Y)$.

You already know that $g$ is bijective. So, it remains to show that $g$ and $g^{-1}$ are continuous.

That $g$ is continuous follows e.g. from the last sentence of Theorem 24.5.

It remains to show that $g^{-1}$ are continuous. By easy manipulations, this reduces to the following:

Claim: Suppose that $0\in Y$, $f(0)=0$, $g(0)=0$, and $Y\ni h\to0$. Then $g^{-1}(h)\to0$.

Proof: Take any real $\de>0$ such that $x\in Y$ whenever $|x|\le\de$, where $|\cdot|$ denotes the Euclidean norm. Since $f$ is strictly convex (and hence continuous on $Y$), there is some real $\ep(\de)>0$ such that $f(x)\ge\ep(\de)$ whenever $|x|=\de$, so that for $\eta(\de):=\frac{\ep(\de)}\de$ we have \begin{equation*} x\in X\ \&\ |x|\ge\de\implies f(x)\ge\eta(\de)\,|x|. \tag{1}\label{1} \end{equation*}

Let $g_h(x):=f(x)-h\cdot x$, where $|h|<\eta(\de)$. Then, by \eqref{1}, \begin{equation*} g_h(x)\ge\Big(\eta(\de)-|h|\Big)|x|>0=g_h(0) \end{equation*} if $x\in X$ and $|x|\ge\de$. So, the continuous strictly convex function $g_h$ attains its minimum on $X$ at a unique point $x_h\in X$ such that $|x_h|<\de$, so that for all $x\in X$ we have $g_h(x)\ge g_h(x_h)$ or, equivalently, $f(x)\ge f(x_h)+h\cdot(x-x_h)$, so that $h=g(x_h)$ and $x_h=g^{-1}(h)$. Thus, $|g^{-1}(h)|<\de$ if $|h|<\eta(\de)$. $\quad\Box$

$\newcommand{\na}{\nabla}\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Let $Y$ denote the interior of $X$. Then indeed $g:=\na f$ is a homeomorphism of $Y$ onto $g(Y)$.

You already know that $g$ is bijective. So, it remains to show that $g$ and $g^{-1}$ are continuous.

That $g$ is continuous follows e.g. from the last sentence of Theorem 24.5.

It remains to show that $g^{-1}$ is continuous. By easy manipulations, this reduces to the following:

Claim: Suppose that $0\in Y$, $f(0)=0$, $g(0)=0$, and $Y\ni h\to0$. Then $g^{-1}(h)\to0$.

Proof: Take any real $\de>0$ such that $x\in Y$ whenever $|x|\le\de$, where $|\cdot|$ denotes the Euclidean norm. Since $f$ is strictly convex (and hence continuous on $Y$), there is some real $\ep(\de)>0$ such that $f(x)\ge\ep(\de)$ whenever $|x|=\de$, so that for $\eta(\de):=\frac{\ep(\de)}\de$ we have \begin{equation*} x\in X\ \&\ |x|\ge\de\implies f(x)\ge\eta(\de)\,|x|. \tag{1}\label{1} \end{equation*}

Let $g_h(x):=f(x)-h\cdot x$, where $|h|<\eta(\de)$. Then, by \eqref{1}, \begin{equation*} g_h(x)\ge\Big(\eta(\de)-|h|\Big)|x|>0=g_h(0) \end{equation*} if $x\in X$ and $|x|\ge\de$. So, the continuous strictly convex function $g_h$ attains its minimum on $X$ at a unique point $x_h\in X$ such that $|x_h|<\de$, so that for all $x\in X$ we have $g_h(x)\ge g_h(x_h)$ or, equivalently, $f(x)\ge f(x_h)+h\cdot(x-x_h)$, so that $h=g(x_h)$ and $x_h=g^{-1}(h)$. Thus, $|g^{-1}(h)|<\de$ if $|h|<\eta(\de)$. $\quad\Box$