After more consideration, I have come up with what I think is an answer to my first question. I won't bore people with all the details, but for me the Aha! moment was when I considered the class $\mathscr{B}_{\Sigma}=\{f\in\mathscr{B}: f\text{ has $\sigma$-compact support}\}$. I can prove $\mathscr{B}_\Sigma$ contains the continuous functions with $\sigma$-compact support and is closed under pointwise limits. I can further prove that $\mathscr{B}\mathscr{B}_\Sigma=\{fg: f\in\mathscr{B},g\in\mathscr{B}_\Sigma\}\subseteq\mathscr{B}_\Sigma$ (which depended on my previously having shown that $\mathscr{B}$ is actually an algebra of functions (in particular, it is closed under multiplication of two functions). Finally, I can prove that every $f\in\mathscr{B}_\Sigma$ is Baire measurable. I'll omit the rest of the proof unless someone asks me to post it.
EDIT
Since the above version doesn't work, let me present, in detail, a version that I think does work.
I'll start with some terminology. Let $\mathbf{F}_0$ represent the class of all continuous real valued functions on $X$. Let $\mathscr{F}$ be the subclass of $\mathbf{F}_0$ consisting of continuous functions whose range is contained in $[0,1]$. If $\mathbf{T}$ is any class of real valued functions on $X$, and $\alpha$ is an ordinal, then we define $\mathscr{B}_\alpha(\mathbf{T})$ by transfinite induction as follows:
Let $\mathscr{B}_0(\mathbf{T})=\mathbf{T}$. If $\alpha$ is a successor ordinal, let $\mathscr{B}_\alpha(\mathbf{T})$ be the class of all pointwise limits of functions in $\mathscr{B}_{\alpha-1}(\mathbf{T})$. If $\alpha\leq\omega_1$ is a limit ordinal (where $\omega_1$ is the first uncountable ordinal), let $\mathscr{B}_\alpha(\mathbf{T})=\bigcup_{\beta<\alpha}\mathscr{B}_\beta(\mathbf{T})$.
Next, we define $\mathscr{B}(\mathbf{T})$ to be the smallest class of real valued functions on $X$ which is closed under pointwise limits and which contains $T$. Thus, the $\mathscr{B}$ of the problem statement is just $\mathscr{B}(\mathbf{F}_0)$.
Proposition 1. $\mathscr{B}(\mathbf{T})=\mathscr{B}_{\omega_1}(\mathbf{T})$.
Proof. Suppose $f_n\in\mathscr{B}_{\omega_1}(\mathbf{T})$ for $n=1,2,\dots$ and that the limit $f=\lim f_n$ exists. Then for each $n=1,2,\dots$, there exists an ordinal $\alpha_n<\omega_1$ such that $f_n\in\mathscr{B}_{\alpha_n}(\mathbf{T})$. Let $\alpha<\omega_1$ be an ordinal greater than each $\alpha_n$. Then $f_n\in\mathscr{B}_\alpha(\mathbf{T})$ for each $n=1,2,\dots$ and $f\in\mathscr{B}_{\alpha+1}(\mathbf{T})\subseteq\mathscr{B}_{\omega_1}(\mathbf{T})$. That is, $\mathscr{B}_{\omega_1}(\mathbf{T})$ is closed under pointwise limits. $\mathbf{T}=\mathscr{B}_0(\mathbf{T})\subseteq\mathscr{B}_{\omega_1}(\mathbf{T})$ so $\mathscr{B}_{\omega_1}(\mathbf{T})$ contains $\mathbf{T}$, hence contains all of $\mathscr{B}(\mathbf{T})$.
Conversely, $\mathscr{B}_{\omega_1}(\mathbf{T})\subseteq\mathscr{B}(\mathbf{T})$, for if not, there exists a function in $\mathscr{B}_{\omega_1}(\mathbf{T})$ which is not in $\mathscr{B}(\mathbf{T})$. Hence the class of countable ordinals $\alpha$ such that $\mathscr{B}_\alpha(\mathbf{T})$ contains a function not in $\mathscr{B}(\mathbf{T})$ is non empty. Let $\alpha$ be the least such ordinal. Then if $\beta<\alpha$, $\mathscr{B}_\beta(\mathbf{T})\subseteq\mathscr{B}(\mathbf{T})$. If $\alpha=0$ we have a contradiction since $\mathscr{B}_0(\mathbf{T})=\mathbf{T}\subseteq\mathscr{B}(\mathbf{T})$. $\alpha$ cannot be a limit ordinal, since then $\mathscr{B}_\beta(\mathbf{T})$ would contain a function not in $\mathscr{B}(\mathbf{T})$ for some $\beta<\alpha$, contradicting the minimality of $\alpha$. So $\alpha=\beta+1$ for some countable ordinal $\beta$, and there is a function $f\in\mathscr{B}_\alpha(\mathbf{T})$ which is not in $\mathscr{B}(\mathbf{T})$. By definition $f=\lim f_n$ for some sequence $\{f_n\}$ of functions in $\mathscr{B}_\beta(\mathbf{T})\subseteq\mathscr{B}(\mathbf{T})$, so by the definition of $\mathscr{B}(\mathbf{T})$, $f\in\mathscr{B}(\mathbf{T})$, a contradiction. $\quad\Box$
Proposition 2. Let $g$ be an arbitrary real valued function on $X$. Let $g\mathscr{B}(\mathbf{T})=\{gf:f\in\mathscr{B}(\mathbf{T})\}$ and let $g\mathbf{T}=\{gf: f\in\mathbf{T}\}$. Then $g\mathscr{B}(\mathbf{T})\subseteq\mathscr{B}(g\mathbf{T})$.
Proof. Suppose not. Then, using Proposition 1, there exists a least countable ordinal $\alpha$ such that $g\mathscr{B}_\alpha(\mathbf{T})$ contains a function not in $\mathscr{B}(g\mathbf{T})$. As in Proposition 1, $\alpha$ cannot be $0$ or a limit ordinal, so let $\alpha=\beta+1$. Then there exists $f\in\mathscr{B}_\alpha(\mathbf{T})$ such that $gf\notin\mathscr{B}(g\mathbf{T})$ and there exists a sequence $\{f_n\}$ in $\mathscr{B}_\beta(\mathbf{T})$ such that $f=\lim f_n$. By the minimality of $\alpha$, $g\mathscr{B}_\beta(\mathbf{T})\subseteq\mathscr{B}(g\mathbf{T})$, so $gf=\lim gf_n\in\mathscr{B}(g\mathbf{T})$, a contradition. $\quad\Box$
Halmos Measure Theory Theorem 50.B. If $C$ is a compact set, $F$ is a closed set, and $C\cap F=\varnothing$, then there exists a function $f$ in $\mathscr{F}$ such that $f(x)=0$ for $x$ in $C$ and $f(x)=1$ for $x$ in $F$.
Halmos Measure Theory Theorem 50.D. If $C$ is compact, $U$ is open, and $C\subseteq U$, then there exist sets $C_0$ and $U_0$ such that $C_0$ is a compact $G_\delta$, $U_0$ is a $\sigma$-compact open set, and
$$C\subseteq U_0\subseteq C_0\subseteq U.$$
Halmos Measure Theory Theorem 51.A. Every Borel set is $\sigma$-bounded; every $\sigma$-bounded open set is a Borel set.
Halmos Measure Theory Theorem 51.B. If a real valued continuous function $f$ on $X$ is such that the set $N(f)=\{x:f(x)\neq 0\}$ is $\sigma$-bounded, then $f$ is Baire measurable.
Proposition 3. Let $g$ be a continuous real valued function on $X$ having $\sigma$-compact support. Then every function in $\mathscr{B}(g\mathbf{F}_0)$ is Baire measurable.
Proof. Let $\mathbf{M}=\{h\in\mathscr{B}(g\mathbf{F}_0):\text{ $h$ is Baire measurable}\}$. $\mathbf{M}$ is closed under pointwise limits since both $\mathscr{B}(g\mathbf{F}_0)$ and the class of Baire measurable functions are. If $f\in\mathbf{F}_0$ then $h=gf$ is continuous and has $\sigma$-compact support, so $N(h)$ is $\sigma$-bounded, so by Theorem 51.B, $h$ is Baire measurable. Hence $g\mathbf{F}_0\subseteq\mathbf{M}$ and therefore $\mathscr{B}(g\mathbf{F}_0)\subseteq\mathbf{M}\subseteq\mathscr{B}(g\mathbf{F}_0)$, so every $h\in\mathscr{B}(g\mathbf{F}_0)$ is Baire measurable. $\quad\Box$.
Proposition 4. If $E$ is a Borel set and if $\chi_E\in\mathscr{B}$ then $E$ is a Baire set.
Proof. By 51.A, $E$ is $\sigma$-bounded, say $E\subseteq\cup C_n$ where $C_n$ is compact, $n=1,2,\dots$. Since a finite union of compact sets is compact, without loss of generality, we may assume $C_1\subseteq C_2\subseteq\cdots$. By 50.D applied to $C_n\subseteq X$, there exist open $U_n$ and comapct $K_n$ such that $C_n\subseteq U_n\subseteq K_n$. By 50.B applied to $C_n$ and $F=X-U_n$, there exists $f_n\in\mathscr{F}$ such that $(1-f_n)(C_n)=\{1\}$ and $(1-f_n)(X-U_n)=\{0\}$. In addition, $(1-f_n)(X-K_n)=\{0\}$ since $X-K_n\subseteq X-U_n$. For $x\in E$, there is an $n_0$ such that $x\in C_n$ for all $n\geq n_0$, so $(1-f_n)(x)\to 1$. For $x\notin\cup K_n$, $x\in X-K_n$, $n=1,2,\dots$, so $(1-f_n)(x)=0$, $n=1,2,\dots$, hence $(1-f_n)(x)\to 0$. Therefore, $g_n=1-f_n$ is continuous with compact support. By Proposition 2, $g_n\mathscr{B}=g_n\mathscr{B}(\mathbf{F}_0)\subseteq\mathscr{B}(g_n\mathbf{F}_0)$. Then, by Proposition 3, since $\chi_E\in\mathscr{B}$, we have that $g_n\chi_E$ is Baire measurable, $n=1,2,\dots$. Since $g_n(x)\to 1$ for $x\in E$, and since $\chi_E(x)=0$ for $x\notin E$, it is clear that $\chi_E=\lim g_n\chi_E$, and hence $\chi_E$ is Baire measurable. For Halmos Measure Theory, which works with $\sigma$-rings,
a function $f$ is measurable if and only if $N(f)\cap f^{-1}(M)$ is measurable for each Borel set $M$ of the real line. Taking $f=\chi_E$ and $M$ to be the entire real line, this says that $E=E\cap X=N(\chi_E)\cap X=N(\chi_E)\cap\chi_E^{-1}(\mathbb{R})$ is measurable; i.e., a Baire set. $\quad\Box$
