If $X_1, X_2$ are two iid Bernoulli random variables, then their sum is the count of successes in two iid Bernoulli experiments. This is the definition of a binomial random variable with parameters $2$ and $p$.
That is all. $\Box$
If you need to demonstrate further:
The probability that there will be exactly $y$ successes among these two variables is determined by measuring the probability of $y$ successes in a row followed by $2-y$ failures, then multiply this by the count of distinct ways to order these results.
That is:
$$\mathsf P(Y=y) = \underline{\qquad\qquad?}$$
Which was to be demonstrated. $\Box$
Alternatively, from first principles we use the Law of Total Probability to show that when we partition the results by the first variable:
$$\begin{align}
\mathsf P(Y=y)
&= \mathsf P(X_1=1, X_2=y-1) + \mathsf P(X_1=0, X_2=y)
\\[1ex]& = p\cdot \mathsf P(X_2=y-1) + (1-p)\cdot \mathsf P(X_2=y)
\\[1ex]& = \begin{cases}
\underline\qquad & : y =0
\\ \underline\qquad & : y=1
\\ \underline\qquad & : y=2
\end{cases}
\end{align}$$
Which was to be demonstrated. $\Box$
Fill in the blanks if required. But really, you don't need to go passed the first tombstone.
You are on the right track with the binomial thinking. I'll just expand on the notation and definitions.
I will refer to the positive values $u, d$ as the up and down values in a coin toss for ease of thinking. An element of $Ω$ then looks like a $T$-tuple with either a $u$ or $d$ in each component, which could be interpreted as the outcome of $T$ consecutive coin tosses (Bernoulli trials).
I suppose $ω_j$ is the $j$-th component projection of the vector $ω = (x_1, \dots, x_T)$:
$$ω_j(ω) = ω_j((x_1, \dots, x_T)) = x_j.$$
The function $P$ is defined for singleton sets of outcomes $ω$, each of which is a list of $T$ possible coin toss outcomes $ω_1(ω), \dots, ω_T(ω)$. Each $ω_j(ω)$ here corresponds to an outcome of a single toss.
The definition of $U(ω)$ just says "count the number of ups in the $T$ tosses". Reading the definition of $P$ and comparing it to the binomial distribution makes it apparent that $p$ is supposed to be the probability of getting an up and $(1-p)$ the one of getting a down in each independent toss, making it a $B(T, p)$ distribution.
An event $A$ will be a collection of outcomes $ω$, each comprising $T$ trials. For example, the event that you get up in the first trial contains several $ω$: all possible combinations of $T$ trials that have up as the first outcome. That would be written like this: $\{ω \in \Omega : ω_1(ω) = u\}.$
The way to compute the probability of an event is given in the definition: just sum all probabilities corresponding to each outcome $ω$. In order to prove that $P$ is a probability it will be necessary to prove that $P(\Omega) = 1$.
Afterwards, the expectation of $ω_1$ can be computed by definition:
$$E_P(ω_1) = \sum_{ω \in \Omega} ω_1(ω) P({ω}) = \sum_{x_1, ..., x_T \in \{u, d\}} x_1 P(\{(x_1, \dots, x_T)\}).$$
The best way I can think of to evaluate this $T$-fold sum is to split it depending on the value of $x_1$:
$$\sum_{x_2, ..., x_T \in \{u, d\}} u \cdot P(\{(u, x_2, \dots, x_T)\}) + \sum_{x_2, ..., x_T \in \{u, d\}} d \cdot P(\{(d, x_2, \dots, x_T)\}).$$
Call these two sums $S_u$ and $S_d$, respectively. Now you can use what you know about $x_1$ in each to evaluate them. For example, for $S_u$ you know that you already have one $u$, so $P(\{(u, x_2, \dots, x_T)\}) = p \cdot p^\alpha\cdot (1-p)^\beta$, where $\alpha$ and $\beta$ are the number of ups and downs in $\{x_2, \dots, x_T\}$.
The $p^\alpha\cdot (1-p)^\beta$ bit is actually the probability mass of a $B(T-1, p)$ distribution for outcomes $\{x_2, \dots, x_T\}$, so summing over all possible combinations of $T-1$ coin toss outcomes should give $1$ (good thing they asked us to prove it first for arbitrary $T$). It follows that $S_u = pu$ and, similarly, $S_d = (1-p)d$.
Best Answer
The answer is no.
Consider the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, where $\Omega = \{0,\ldots,n\}$, $\mathcal{F} = \mathcal{P}(\Omega)$ and $\mathbb{P}: \mathcal{F} \to [0,1]$ defined as $\mathbb{P}(A) = \sum\limits_{i \in A} \binom ni p^i(1-p)^{n-i}$ for every $A \in \mathcal{F}$.
Consider now $X: \Omega \to \{0,\ldots, n\}$ as $X(i)=i$ (the identity function).
The distribution of $X$ is clearly binomial.
Suppose now that $X = X_1 + \ldots + X_n$ with $X_i$ iid with law Bernoulli of parameter $p$, all defined on $(\Omega, \mathcal{F}, \mathbb{P})$.
This would mean that there is $A \in \mathcal{F}$ such that $\mathbb{P}(A) = p$.
And this must be true for every $p \in (0,1)$.
Consider the map $F: (0,1) \to \mathcal{P}(\{0,\ldots,n\})$ that for a given $p$ returns such a set $A$. Since the domain is infinite and the codomain is finite there must be a given $A \subseteq \{0,\ldots,n\}$ such that for infinite many $p$ we have $$\sum\limits_{i \in A} \binom ni p^i(1-p)^{n-i} = p$$ Therefore we have $\sum\limits_{i \in A} \binom ni x^i(1-x)^{n-i} = x$ for every real $x$ (polynomial identity).
And now, for example with $n=2$, we obtain that the LHS can be equal to $0,1,x^2,(1-x)^2,2x(1-x),1-x^2,1-(1-x)^2,1-2x(1-x)$, none of which is equal to $x$.