We have the following setting; let $\{X_{i}:i\in\mathbb{N}\}$ be a sequence of independent identically distributed random variables representing the outcomes of our coin tosses, i.e. taking values $1$ and $-1$. We have that $\mathbb{P}(X_{i}=1)=p$ and $\mathbb{P}(X_{i}=-1)=1-p$.
Let $x\in\mathbb{N}$ be an arbitrary integer. I want to prove that we throw $x$ heads in a row, in finite time.
Define the following sequence $\{Y_{i}:i\in\mathbb{N}\}$ with: $Y_{i}=\left\{\begin{matrix} 1 \quad , \quad \sum^{x+i}_{j=i}X_{j}=x\\ 0 \quad , \qquad\qquad \quad else\\ \end{matrix}\right.$
We then know for the expected value of $Y_{i}$ that: $\mathbb{E}[Y_{i}]=1*p^{x}+0*(1-p^{x})=p^{x}$
Now using the law of large numbers we find that: $\lim_{n\rightarrow\infty}(\frac{1}{n}\sum^{n}_{i=1}Y_{i})=\mathbb{E}[Y_{i}]$ from which we can conclude that: $\lim_{n\rightarrow\infty}(\sum^{n}_{i=1}Y_{i})=\infty$. This would prove that we definitely throw $x$ heads in a row at some point in time.
My question is; is this proof sufficient to say that we throw $x$ heads in finite time? If so, why? We say that it happens eventually when we start throwing infinitely many times, so this isn't saying it happens in a finite amount of throws, right?
I'm not sure I understand what you're concerned about.
The (Strong) Law of Large Numbers implies that with probability $1$, $Y_i = 1$ infinitely many times. Each time $Y_i = 1$ is "in a finite amount of throws". Of course you can't say in advance how many throws this will be.