A possible choice of space $\Omega$ to define Brownian motion is $\Omega=C(\mathbb R_+,\mathbb R)$, then the Brownian motion $(B_t)_{t\in\mathbb R_+}$ is simply the coordinate process, that is, for every $t$ in $\mathbb R_+$ and $\omega$ in $\Omega$, $B_t(\omega)=\omega(t)$.
In this construction, sample paths are the elements $\omega$ of $\Omega$.
But, as is usual in probability, one may prefer not to specify $\Omega$. Then $\Omega$ can be any space large enough for a family $(X_t)_{t\in\mathbb R_+}$ of random variables with the prescribed properties to exist on $\Omega$. Then a sample path is, for some $\omega$ in $\Omega$, the function $X(\omega):\mathbb R_+\to\mathbb R$, $t\mapsto X_t(\omega)$.
Lévy's construction by dichotomy, which you recall, might then correspond to $\Omega=S^\mathbb N$ the product of a countable number of copies of a probability space $(S,\mathcal S,Q)$ being large enough for one standard normal random variable $\xi$ to be defined on it. Then a Brownian motion $(X_t)_{t\in\mathbb R_+}$ on $\Omega$ can be defined as Lévy indicated using the i.i.d. copies of $\xi$ defined on each factor of $\Omega$. Thus, every $\omega$ in $\Omega$ is $\omega=(s_n)_{n\in\mathbb N}$ for some $s_n$ in $S$, and the random variables $X_1$ and $X_{1/2}$, say, are defined by $X_{1}(\omega)=\xi(s_1)$ and $X_{1/2}(\omega)=\frac12\xi(s_1)+\frac12\xi(s_2)$.
Edit: Recall that the $n$th approximation $X^{(n)}$ of the Brownian motion $X$ on $[0,1]$ is piecewise linear on each interval $[(k-1)/2^n,k/2^n]$ with $1\leqslant k\leqslant2^n$. Thus, the $0$th approximation is such that $X^{(0)}_t(\omega)=t\xi(s_1)$ for every $t$ in $[0,1]$, after that, $X^{(n+1)}_t=X^{(n)}_t$ at every $t=k/2^n$ and $2X^{(n+1)}_t=X^{(n)}_{k/2^n}+X^{(n)}_{(k+1)/2^n}+\xi(s_*)/2^{n/2}$ at every $t=(2k+1)/2^{n+1}$, where $*$ is the first index $i$ such that $\xi(s_i)$ is not used yet.
This will follow from the definition if the joint distribution of set of normally distributed random variables(not nececarrily independent) are jointly multivarite distributed?
Yes.
But does this mean that any linear combination of normally distributed random variables are normally distributed, even if they are not independent?
No. A frequently mentioned counterexample is based on $X$ standard normal and $Y=SX$ with $S=\pm1$ symmetric and independent of $X$. Then $X$ and $Y$ are normal but $X+Y$ is not since $P(X+Y=0)=\frac12$, a property that no normal random variable satisfies.
Best Answer
You can choose $\lambda_q,\dots,\lambda_k$ so that $$Y = a_1B_1 + \dots + a_kB_k = \lambda_1B_1 + \lambda_2(B_2-B_1) + \dots + \lambda_k(B_k-B_{k-1}).$$ But from the definition of Brownian motion, you know that $B_1, B_2-B_1, \dots, B_k-B_{k-1}$ are normally distributed and independent, so a linear combination of them is again normally distributed.