Consider an ambient metric space $(\mathcal{X},\Vert\cdot\Vert_\infty)$. Let $\mathcal{B}_1 = \mathcal{B}_{\Vert\cdot\Vert_K}(0,1)\subseteq\mathcal{X}$ be the closed unit ball with respect to some norm $\Vert\cdot\Vert_K$. Denote the $\varepsilon$-covering number of $\mathcal{B}_1$ with respect to $\Vert\cdot\Vert_\infty$ by $\mathcal{N}(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert_\infty)$. That is, we can find a set of points $\{x_1,\dots,x_n\}\subseteq\mathcal{X}$ with $n = \mathcal{N}\left(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert_\infty\right)$ such that for all $x\in\mathcal{B}_1$, there exists $i\in[n]$ with
\begin{equation}
\Vert x - x_i\Vert_\infty\leq\varepsilon
\end{equation}
From the observation that the ball is closed and the covering number is defined with $\leq$ instead of $<$ (see equation above), I am tempted to assume that the covering number is a **right-continuous** function of $\varepsilon$. Is this assumption correct?

------------------------------------------------------------------

Following the answer of @Fedor Petrov, we assume that $(\mathcal{X},\Vert\cdot\Vert_\infty)$ is a *compact* metric space.

**Proof Attempt**

To prove right-continuity of
\begin{equation}
\varepsilon\mapsto\mathcal{N}(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert)
\end{equation}
on $\mathbb{R}_+$, fix some $\varepsilon > 0$. We will show that for every decreasing sequence of positive numbers $(\varepsilon_k)_{k\in\mathbb{N}}$ such that $\varepsilon_k \downarrow\varepsilon$, we have
\begin{equation}
\lim_{k\to\infty} \mathcal{N}(\varepsilon_k, \mathcal{B}_1,\Vert\cdot\Vert) = \mathcal{N}(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert)
\end{equation}
For $k\in\mathbb{N}$, let $n^{(k)} = \mathcal{N}(\varepsilon_k, \mathcal{B}_1,\Vert\cdot\Vert)$. If $\{x_1^{(k)},\dots,x_{n^{(k)}}^{(k)}\}$ is a minimal $\varepsilon_k$-covering of $\mathcal{B}_1$ with respect to $\Vert\cdot\Vert$, then, for all $x\in\mathcal{B}_1$, there exists $i\in[n^{(k)}]$ with
\begin{equation}
\left\Vert x - x_i^{(k)}\right\Vert_\infty\leq\varepsilon_k
\end{equation}
As $k\to\infty$, we have that, for all $x\in\mathcal{B}_1$, there exists $i\in[\lim_{k\to\infty}n^{(k)}]$ with
\begin{equation}
\lim_{k\to\infty} \left\Vert x - x_i^{(k)}\right\Vert_\infty\leq\varepsilon
\end{equation}
Therefore, it immediately follows that
\begin{equation}
\mathcal{N}(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert)\leq\lim_{k\to\infty} \mathcal{N}(\varepsilon_k, \mathcal{B}_1,\Vert\cdot\Vert)
\end{equation}
From the basic properties of the covering number, we have
\begin{equation}
\mathcal{N}(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert)\geq\mathcal{N}(\varepsilon_k, \mathcal{B}_1,\Vert\cdot\Vert)
\end{equation}
for all $k\in\mathbb{N}$ since $\varepsilon\leq\varepsilon_k$. This establishes the claim, specifically, we have shown that
\begin{equation}
\lim_{k\to\infty} \mathcal{N}(\varepsilon_k, \mathcal{B}_1,\Vert\cdot\Vert) = \mathcal{N}(\varepsilon, \mathcal{B}_1,\Vert\cdot\Vert)
\end{equation}
Since $\varepsilon$ was chosen arbitrarily on $\mathbb{R}_+$, we deduce that the covering number is right-continuous on $\mathbb{R}_+$.