Bounding Bernoulli trials by the standard Bernoulli process Suppose we have a Bernoulli-like process $P$. At each step a coin is tossed and the outcome ("success", "failure") is recorded. What differentiate $P$ from the standard Bernoulli process, is that we pick a probability of "success" uniformly at random in range $(1/2, 1)$ at each step before we toss the coin.

I'm interested in finding an upper bound on the expected number of trials 
  until the first "success" is tossed.

What I have thought, if the probability of "success" is at least $1/2$, then at each step $P$ is more probable to stop than a standard Bernoulli process, therefore an expectation of a standard geometrically distributed variable bounds from above the expectation of steps until the first "success".

How can I make this claim formal?

 A: As  David K states implies, your process is exactly a Bernoulli process with non-random success probability $p=3/4$. The expected number of flips is then $4/3\approx1.333$.
Your argument & approach are good.  You can* construct an iid sequence $U_i$ of $U[0,1]$ variables and another, $S_i$,  iid $U[1/2,1]$, and consider the sequence of coupled binary outcomes $(X_i,Y_i)$ where $X_i = 1$ exactly when $U_i\le 1/2$ and $Y_i = 1$ exactly when $U_i\le S_i$.  Then the $X_i$ process has the same probability distribution as the standard Bernoulli process and the $Y_i$ process has the same probability distribution as your $P$ process, and $X_i\le Y_i$ with probability $1$.
Footnote: If you are afraid your  original probability space $(\Omega,\mathcal A, P)$ is not rich enough to support all these newly constructed rvs, don't worry.  It is rich enough to support a $U[1/2,1]$ random variable, and hence is a so-called standard probability space.  If it supports a uniform rv, that rv's binary digits are an iid sequence of fair coin flips, and by Cantor, a countable sequence of such sequences, and thus a countable sequence of uniforms, and so on.  The resulting $X_i$ and $Y_i$ constructed this way will not be equal $\omega$ by $\omega$ to what you started out with, but will have the same distributional properties.
A: While, kimchi's answer is an answer to the problem as I have stated it in the first place ...
I want to share an approach, that tackles directly the expectation bound.
Suppose, we have a series of independent Bernoulli trials $X_i$ each with a probability of success $p_i \geq \frac{1}{2}$. And a series of standard i.i.d. Bernoulli trials $Y_i$ with a probability of success $p = \frac{1}{2}$.
Define by $X$ - the index of the first success in the series $X_i$, and by $Y$ - an index of the first success in the series $Y_i$.
We have that $Y \sim Geom(\frac{1}{2})$, and $\mathbb{E}(Y) = 2$.
We ask, what is $\mathbb{P}(X > k)$ ?
In other words what is the probability that the first success in series $X_i$ happens after the $k^{th}$ trial. The answer can be calculated in a straight-forward way:
$$\mathbb{P}(X > k) = \prod\limits_{i=1}^k(1-p_i),$$ as all trials before and including $k^{th}$ should fail.
We further use the fact that $p_i \geq p$ to show that
$$\mathbb{P}(X > k) = \prod\limits_{i=1}^k(1-p_i) \leq \prod\limits_{i=1}^k(1-p) = \mathbb{P}(Y > k)$$
Recall, that for a discrete variable $Z$(such as $X$ and $Y$) taking values in $\{1, 2, \ldots \} \cup \{ +\infty\}$
$$\mathbb{E}(Z) = \sum\limits_{k = 1}^{\infty}\mathbb{P}(Z > k)$$
Now sum the probabilities to get the expectations:
$$\mathbb{E}(X) = \sum\limits_{k = 1}^{\infty}\mathbb{P}(X > k) \leq \sum\limits_{k = 1}^{\infty}\mathbb{P}(Y > k) = \mathbb{E}(Y)$$
Thus a direct upper bound is shown.
