The answer is no.

**Lemma:** Every quasi-totally bounded (instead of finite, there is a countable net) metric space $\displaystyle M$ is separable.

**Proof:** For every $\displaystyle n\in\mathbb{N}$ let $\displaystyle \mathcal{A}_n$ be the guaranteed countable $\displaystyle \tfrac{1}{n}$-net and let $\displaystyle \mathcal{A}=\bigcup_{n=1}^{\infty}\mathcal{A}_n$. Clearly $\displaystyle \mathcal{A}$ being the countable union of countable sets is countable. Also, we claim that it is dense in $\displaystyle M$. To see this let $\displaystyle m\in M$ and $\displaystyle \varepsilon>0$ be given. By the Archimedean principle there exist some $\displaystyle n\in\mathbb{N}$ such that $\displaystyle \frac{1}{n}<\varepsilon$. So, then since $\displaystyle \left\{B_{\frac{1}{n}}(a)\right\}_{a\in\mathcal{A} _n}$ cover $\displaystyle M$ it must be that $\displaystyle x\in B_{\frac{1}{n}}(a)$ for some $\displaystyle a\in\mathcal{A}_n\subseteq\mathcal{A}$. Thus, for that $\displaystyle a$ we have that $\displaystyle d(x,a)<\frac{1}{n}<\varepsilon$ from where the conclusion follows. $\displaystyle \blacksquare$

It follows since every compact metric space is totally bounded that every compact metric space is separable.

Also:

**Lemma:** Let $\displaystyle X$ be a $\displaystyle T_1$ space with $\displaystyle \mathfrak{D}\subseteq X$ dense. Then, $\displaystyle \text{card }X\leqslant 2^{2^{\text{card }\mathfrak{D}}}=2\text{ card }\mathcal{P}\left(\mathcal{P}\left(\mathfrak{D}\ri ght)\right)$

**Proof:** For a fixed $\displaystyle x\in X$ define $\displaystyle \mathfrak{N}_x$ to be the set of all neighborhoods of $\displaystyle x$ and define $\displaystyle \Upsilon:X\to\mathcal{P}\left(\mathcal{P}\left(\ma thfrak{D}\right)\right):x\mapsto\left\{N\cap\mathf rak{D}:N\in\mathfrak{N}_x\right\}$. To see this is an injection we merely note that if $\displaystyle x\ne y$ then $\displaystyle X$ being $\displaystyle T_1$ guarantees there are neighboorhoods $\displaystyle U\in\mathfrak{N}_x$ and $\displaystyle V\in\mathfrak{N}_y$ such that $\displaystyle y\notin U$ and $\displaystyle x\notin V$. In particular since every neighborhood of $\displaystyle y$ contains $\displaystyle y$ we have that $\displaystyle U\notin\mathfrak{N}_x$ and so $\displaystyle U\cap N\notin\Upsilon(y)$ and so $\displaystyle \Upsilon(x)\ne\Upsilon(y)$. It follows that $\displaystyle \Upsilon$ is an injection from where the conclusion follows. $\displaystyle \blacksquare$

Thus, either looking

here, or noticing that $\displaystyle \text{card }\mathbb{R}^{\mathbb{R}}=\mathfrak{c}^{\mathfrak{c }=\left(2^{\aleph_0}\right)^{\mathfrak{c}}=2^{\ale ph_0}\cdot\mathfrak{c}}=2^{\mathfrak{c}}=\text{car d }\mathcal{P}(\mathbb{R})=\text{card }\mathcal{P}\left(\mathcal{P}\left(\mathbb{N}\righ t)\right)$ and so $\displaystyle \text{card }\mathcal{P}\left(\mathbb{R}^{\mathbb{R}}\right)=\ text{card }\mathcal{P}\left(\mathcal{P}\left(\mathcal{P}\lef t(\mathbb{N}\right)\right)\right)>\text{card }\mathcal{P}\left(\mathcal{P}\left(\mathbb{N}\righ t)\right)$ and so by the above lemma $\displaystyle \mathcal{P}\left(\mathbb{R}^{\mathbb{R}}\right)$ cannot be turned into a separable $\displaystyle T_1$ space and thus not into a compact metric space.