Ask the proof for bounded and finite-dimensional implies totally bounded?

The definition of totally bounded is: A metric space $(\mathbb{M}, d)$ is totally bounded if for any $\epsilon>0$ there exists $y_1, \ldots, y_n$ in $\mathbb{M}$ such that $\mathbb{M} \subset \cup_{i=1}^n\left\{x \in \mathbb{M}: d\left(x, y_i\right)<\epsilon\right\}$ for some finite positive integer $n$.

Now given a space $X$ which is bounded and finite-dimensional, how to show it is totally bounded?

The background of the question is from the proof of a theorem:

Theorem Let $f$ be a measurable function from $E$ to $\mathbb{X}$ with

$$\int_E\|f\| d \mu<\infty$$

Suppose that for each $n$ there exists a finite-dimensional subspace $\mathbb{X}_n$ of $\mathbb{X}$ such that

$$\lim _{n \rightarrow \infty} \int_E\left\|f-g_n\right\| d \mu=0$$

for some measurable $g_n$ taking value in $\mathbb{X}_n$. Then, there exist simple and Bochner integrable functions $f_n$ such that $\lim _{n \rightarrow \infty} \int_E\left\|f_n-f\right\| d \mu=0$ holds.

Proof: Define

$$\tilde{\mathbb{X}}_n=\mathbb{X}_n \cap\left\{g \in \mathbb{X}:\|g\| \in\left[n^{-1}, n\right]\right\}$$

and

$$E_n=\left\{\omega \in E: g_n(\omega) \in \tilde{\mathbb{X}}_n\right\}$$

Markov's inequality produces

$$\mu\left(E_n\right) \leq n \int_{E_n}\left\|g_n\right\| d \mu \leq n \int_E\left\|g_n\right\| d \mu<\infty$$

As $\tilde{\mathbb{X}}_n$is bounded and finite-dimensional, it is totally bounded. Thus, the Heine-Borel theorem (Theorem 2.1.17: A metric space is compact if it is complete and totally bounded.) can be invoked to see that there is a finite partition $B_i, 1 \leq i \leq k$, of $\tilde{\mathbb{X}}_n$ such that each $B_i$ is in the Borel $\sigma$-field for $\mathbb{X}$ and has diameter less than $\left(n \mu\left(E_n\right)\right)^{-1}$. For an arbitrary element $b_i \in B_i$, set

$$f_n(\omega)=\sum_{i=1}^k b_i I_{\left\{g_n(\omega) \in B_i\right\}}$$

Note that $f_n=0$ on $E_n^c$ and hence (2.24) entails that $f_n$ is simple and Bochner integrable as a simple function. Also, by construction,

$$\left\|f_n(\omega)-g_n(\omega)\right\| \leq \max _{1 \leq i \leq k} \sup _{x \in B_i}\left\|b_i-x\right\| \leq\left(n \mu\left(E_n\right)\right)^{-1}$$

for $\omega \in E_n$. Thus, by the triangle inequality,

$$\begin{aligned}& \int_E\left\|f_n-f\right\| d \mu \\& \quad \leq \int_E\left\|f_n-g_n\right\| d \mu+\int_E\left\|g_n-f\right\| d \mu \\& \quad=\int_{E_n}\left\|f_n-g_n\right\| d \mu+\int_{E_n^c}\left\|f_n-g_n\right\| d \mu+\int_E\left\|g_n-f\right\| d \mu\end{aligned}$$

The first and third terms in the last expression tend to zero by (2.25) and (2.23), respectively, while the second term reduces to $\int_{E_n^c}\left\|g_n\right\| d \mu$ because $f_n=$ 0 on $E_n^c$. Thus, to verify (2.19), we only need to establish that

$$\lim _{n \rightarrow \infty} \int_{E_n^c}\left\|g_n\right\| d \mu=0$$

or, equivalently, by (2.23) that

$$\lim _{n \rightarrow \infty} \int_E\|f\| I\left(\left\|g_n\right\|>n\right) d \mu=0$$

and that

$$\lim _{n \rightarrow \infty} \int_E\|f\| I\left(\left\|g_n\right\|<n^{-1}\right) d \mu=0$$

First, from (2.23) and Markov's inequality,

$$\mu\left(\left\|g_n\right\|>n\right) \leq n^{-1} \int_E\left\|g_n\right\| d \mu \rightarrow 0$$

As $\|f\|$ is integrable, (2.26) follows easily (cf. Exercise 5.6 of Resnick, 1998) from this relation. To show (2.27), for $\epsilon>0$, write

$$\begin{aligned}\int_E\|f\| I\left(\left\|g_n\right\|<n^{-1}\right) d \mu= & \int_E\|f\| I\left(\left\|g_n\right\|<n^{-1},\|f\|>\epsilon\right) d \mu \\& +\int_E\|f\| I\left(\left\|g_n\right\|<n^{-1},\|f\| \leq \epsilon\right) d \mu\end{aligned}$$

so that

$$\begin{aligned}& \limsup _{n \rightarrow \infty} \int_E\|f\| I\left(\left\|g_n\right\|<n^{-1}\right) d \mu \leq \limsup _{n \rightarrow \infty} \int_E\|f\| I\left(\left\|g_n\right\|<n^{-1},\|f\|>\epsilon\right) d \mu \\& +\int_E\|f\| I(\|f\| \leq \epsilon) d \mu .\end{aligned}$$

The first term on the right of the inequality is zero as

$$\begin{aligned}\mu\left(\left\|g_n\right\|<n^{-1},\|f\|>\epsilon\right) & \leq \frac{\int_E\left\|f-g_n\right\| I\left(\left\|g_n\right\|<n^{-1},\|f\|>\epsilon\right) d \mu}{\epsilon-n^{-1}} \\& \leq \frac{\int_E\left\|f-g_n\right\| d \mu}{\epsilon-n^{-1}} \rightarrow 0\end{aligned}$$

due to Markov's inequality and (2.23). Hence, (2.27) follows by now letting $\epsilon \rightarrow 0$ and applying Lebesgue's dominated convergence theorem.