First, a martingale is always only specified with respect to a filtration, and so is thus a local martingale. You do not specify any filtration in your problem, so I assume you mean the natural filtration of the local martingale (i.e., the smallest filtration w.r.t. which $X$ is a local martingale).
Second, your prove/disprove statement does not consist of one, but of multiple claims. I will try to disentangle them first before discussing their correctness. I will refer throughout to the the text Continuous Martingales and Brownian Motion which you mention, too.
(1) Let $X$ be a continuous local martingale, then there exists a Brownian motion $B$ and a predictable process $\xi$ such that $X_t = B_{\int_0^t \xi_s^2 \, ds}$ for all $t \in [0, \infty)$.
(2) Let $X$ be a continuous local martingale, then there exists a Brownian motion $\tilde{B}$ and a predictable process $\tilde{\xi}$ such that $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$ for all $t \in [0, \infty)$.
(3) $\xi$ and $\tilde{\xi}$ as well as $B$ and $\tilde{B}$ agree in an appropriate sense, say, are versions of each other.
(4) The quadratic variation of $\langle X \rangle_t$ is then given by $\int_0^t \xi^2_s \, ds$ (resp. $\int_0^t \tilde{\xi}^2_s \, ds$).
(5) It holds true that $\xi$ is increasing and of finite variation.
(6) The statement (3) holds true at least in the case when $\xi$ is is deterministic.
Lets now asses the claims in detail:
(1) This statement holds true as long as $\langle X \rangle_\infty = \infty$, it is called Dambis -- Dubins-Schwarz theorem (cf. R-Y, Thm. V.1.6). Without the assumption this is not necessarily true in the natural filtration (as it might be not rich enough to support a Brownian motion). However, it is possible to enlarge the probability space resp. the filtration such that this statement is true in the larger space (cf. R-Y, Thm V.1.7).
(2) This is the \textit{martingale representation theorem}. It is usually formulated in the context of Brownian filtrations, i.e., it is a priori assumed that $X$ is a local martingale with respect to a filtration generated by a Brownian motion. However, it holds true without this assumption as long as the measure generated by the quadratic variation process, $d\langle X \rangle_t$, is equivalent to the Lebesgue measure $dt$. If there is no equivalence between the measures but at least $d\langle X\rangle_t$ is absolutely continuous with respect to the Lebesgue measure, one can save again the situation by enlarging the probability space (cf. R-Y, V.3.8 and discussion thereafter).
Without absolute continuity the statement is false as can be seen on the following example. Let $C_t^i$ be copies of the Cantor function on the unit interval $[0,1)$ and define the extension to the positive halfline by
$$ K_t = \sum_{i=0}^\infty i + C_{t-i}^i {{\mathchoice{1\mskip-4mu\mathrm l}{1\mskip-4mu\mathrm l}
{1\mskip-4.5mu\mathrm l}{1\mskip-5mu\mathrm l}}}_{[i,i+1)}(t) $$
Let $W$ be a Brownian motion and define $X$ by $X_t = W_{K_t}$. This is clearly a martingale with quadratic variation $\langle X \rangle_t = K_t$. However, if $X$ could be written $X_t = \int_0^t \tilde{\xi}_s \, d\tilde{B}_s$, then we would have by (4) $\langle X \rangle_t = \int_0^t \tilde{\xi}^2_s \, ds$. However, this is always an absolutely continuous function whereas $K$ is not, therefore they cannot agree.
(3) From the restrictions above it is clear that this cannot hold true in general. But you can even find very simple counterexamples, e.g., set $\tilde{\xi} =1$. Then you have
$$ \int_0^t (-1) \, d\tilde{B}_s = - \tilde{B}_t = B_t = B_{\int_0^t 1^2 \, ds} = B_{\int_0^t (-1)^2 \, ds}$$
So in this case $\tilde{B} = - B$ whereas $\xi$ is not uniquely determined (e.g., it could be $1$ or $-1$). This can be of course made worse using processes jumping between $1$ and $-1$.
(4) This is true and in both cases (direct calculation using properties of Brownian motion and R-Y, Prop. IV.2.7 resp.).
(5) Every increasing process is of finite variation by definition. However, the claimed result is clearly not true, just think about an arbitrary integral with deterministic non- monotone integrand, i.e. $X_t = \int_0^t \sin{(s)} \, \, dW_s$ for some Brownian motion $W$. I assume this is simply a typo and you intended to claim that the quadratic variation itself is increasing (or rather: non-decreasing) and thus of bounded variation (which is evidently true, you just add more squares).
(6) No, the example from (3) or the Cantor function example from (2) are counterexamples.
Let me conclude with two remarks about the context:
(a) The idea to place stochastic calculus on change of time instead of stochastic integration has some interesting history. Actually, it predates Ito calculus and was the first form of stochastic calculus, developed by Wolfgang Döblin (Vincent Doblin) in his famous lettre scellée to the French Academy of Science, https://mathoverflow.net/a/100040/20026.
(b) I disagree with your statement that local martingale is a misnomer. Yes, a local martingale is much more general than a martingale. But it is intentionally not called generalized martingale but local martingale, as it is locally (i.e. up to an increasing sequence of stopping times) a martingales. Generalizations that resemble locally the original object are very often far more general than the original object. The gap between manifolds and Euclidean spaces is for sure much larger than that between true and local martingales.
Best Answer
$W^a$ is in fact a martingale. To see this, write $W^a(t) = W(t \land T_a)$. See also Theorem 3.39 here.
When you write an expression like $\mathbb{E}(W^a(T^a))$ you are implicitly assuming that $W^a(T^a)$ is measurable. This requires $t \ge T_a$ (and trivializes the expectation).