Here is a full answer to the question in the title:
The conditional expectation of a stochastic process is a martingale if the following conditions are satisfied :
- $X_t$ is $\mathcal{F}_t$-adapted.
- $\forall t \geq 0$, $X_t$ is integrable.
- $\forall t \geq s \geq 0, \operatorname{E}[X_t \mid \mathcal{F}_s]= X_s$ a.s.
Let's note $\forall 0 \leq t \leq T, X_{t} =\operatorname{E}[Y_{T} \mid \mathcal{F}_{t}]$
For the first condition, as @saz's comment says, by the definition of the conditional expectation, $\operatorname{E}[Y_T \mid \mathcal{F}_t]$ is $\mathcal{F}_t$-adapted for any $t$.
For the second condition, we have to show that $\operatorname{E}[\vert \operatorname{E}[Y_{T} \mid \mathcal{F}_{t}] \vert]<+\infty$, which is the case because $\mathcal{F}_{t}$ is a filtration generated by a random variable on a finite time and $Y$ only has finite values.
The third condition can be proven with the law of total expectations:
Let $(\Omega,\mathcal{F}_t,\operatorname{P})$ be a probability space, $\mathcal{F}_t$ being the filtration generated by $Y_t$. For a random variable $Y_t$ on such a space, the law of total expectations states that if $\operatorname{E}[Y_t]$ is defined (e.g. $\operatorname{E}[\vert Y_t \vert]<+\infty$), and with $s \leq t \leq T$, we have:
$\operatorname{E}[ \operatorname{E}[Y_T \mid \mathcal{F}_t] \mid \mathcal{F}_s] = \operatorname{E}[Y_T \mid \mathcal{F}_s]$
Which is the same as:
$\operatorname{E}[X_t \mid \mathcal{F}_s] = X_s$, and we then have the 3rd condition.
In short, the answer to the question in the title is yes in most cases that you will encounter in exercises or job interviews because they are almost always designed to be solved.
Best Answer
First of all, we notice that \begin{align} E[(X_t - X_s)(Y_t - Y_s)| \mathcal F_s] &= E[X_tY_t - X_tY_s -X_sY_t +\underbrace{X_sY_s}_{\mathcal{F}_s-\text{mesurable}}|\mathcal F_s]\\ &= E[X_tY_t|\mathcal F_s] - Y_sE[X_t|\mathcal F_s] - X_sE[Y_t|\mathcal F_s] + X_sY_s \\ &= E[X_tY_t|\mathcal F_s] - Y_sX_s - X_sY_s + X_sY_s \\ &=E[X_tY_t-X_sY_s|\mathcal F_s]\tag{1} \end{align} The third equality comes from the fact that $X$ and $Y$ are martingales.
Let's do the $\Rightarrow$ first: Assume that $\lbrace{Z_t\rbrace}_{t\geq0}$ is a martingale. We have: for $s \leq t$ \begin{align} E[(X_t - X_s)(Y_t - Y_s)| \mathcal F_s] &= E[X_tY_t - X_sY_s|\mathcal F_s]\\ &= E[X_tY_t |\mathcal F_s]- X_sY_s \\ &= X_sY_s- X_sY_s = 0 \end{align} The last equality follows from the martingale property.
Now the $\Leftarrow$. We assume that $E[(X_t - X_s)(Y_t - Y_s)| \mathcal F_s] =0$. Using $(1)$, we have : \begin{align} &E[X_tY_t - X_sY_s|\mathcal F_s] =0\\ &\Rightarrow\quad E[X_tY_t|\mathcal F_s] = X_sY_s \end{align} Plus, we know that $Z\in L_1$. Therefore, we can conclude that $Z$ is a martingale.