Since $X$ is second countable, all you need for $X^*$ to be second countable is a countable neighborhood base for the point at infinity, right? Since $X$ is locally compact and second countable, you can cover $X$ with countably many open sets whose closures are compact. Then every compact subset of $X$ is contained in a finite union of those open sets, and the complements of the closures of those unions will provide a countable neighborhood base for $\infty$.
You don't need any assumptions to show that for every topological space $X$, $C_0(X)$ is a closed subspace of $C_b(X)$. Since $C_b(X)$ is a Banach space, endowed with the supremum norm, it suffices to show that the limit of every uniformly convergent sequence of functions vanishing at infinity also vanishes at infinity.
So let $(f_n)$ be a sequence in $C_0(X)$, converging uniformly to $f \in C_b(X)$. Let $\varepsilon > 0$ be given. By the uniform convergence, there is an $N$ such that $\lVert f_N - f\rVert_{\infty} < \varepsilon/2$. Since $f_N \in C_0(X)$,
$$K_{N,\varepsilon/2} := f_N^{-1}(\{ z : \lvert z\rvert \geqslant \varepsilon/2\})$$
is a closed (quasi)compact subset of $X$. Then
$$M_{\varepsilon} := f^{-1}(\{ z : \lvert z\rvert \geqslant \varepsilon\})$$
is a closed - by continuity of $f$ - subset of $K_{N,\varepsilon/2}$, and hence (quasi)compact.
Since $\varepsilon > 0$, it follows that $f \in C_0(X)$.
However, on many spaces, $C_0(X)$ is a rather boring space - namely the trivial vector space. For if there is an $f \in C_0(X) \setminus \{0\}$, then for small enough $\varepsilon > 0$ the set $f^{-1}(\{ z : \lvert z\rvert \geqslant \varepsilon\})$ is a compact subset of $X$ with nonempty interior. In infinite-dimensional Hausdorff topological vector spaces, no compact subset has nonempty interior. So $C_0(L_2(\Omega)) = \{0\}$ (unless $L_2(\Omega)$ is finite-dimensional, which happens for certain measures).
Best Answer
First of all, a compact metric space is second countable, hence a metrizable space can be homeomorphic to a subspace of a compact metrizable space only if it is second countable.
So let's assume $X$ is a separable metrizable space. Choose a compatible metric $0 \leq d \leq 1$, and complete $X$ to get a Polish space $\overline{X}$ with a homeomorphic copy of $X$ inside. Now, as I argued in my answer here, every Polish space is homeomorphic to a $G_{\delta}$ inside the Hilbert cube $[0,1]^{\mathbb{N}}$.
The embedding itself is easy, simply choose a dense subset $(x_n)_{n \in \mathbb{N}} \subset X$ and map $x$ to $(d(x,x_n))_{n \in \mathbb{N}} \in [0,1]^{\mathbb{N}}$ (recall that we chose a bounded metric $0 \leq d \leq 1$). This is obviously continuous, and it is not hard to show that it's a homeomorphism onto its image. To see that the image of $\overline{X}$ is a $G_{\delta}$ is harder and given in detail in the answer to Apostolos's question I mentioned above.
Upshot: every separable metrizable space is homeomorphic to a subspace of the Hilbert cube (this one of the 100 variants and refinements of the Urysohn theorem).
Note:
Finally, if a locally compact space is metrizable, then it is necessarily second countable, hence its one-point compactification is second countable as well, and, again by Urysohn metrizable and $\overline{X}$ is an open subset of its one-point compactification.
For more on this, consult Kechris, Classical descriptive set theory, Springer GTM 156, Springer 1994. See in particular Theorem 5.3 on page 29.