Let us suppose towards contradiction that you play according to a strategy that allows play to continue for $\omega_1$ many stages. At each countable stage $\alpha$, you have countably many coins $A_\alpha$ in your possession. One of them will be played right at that stage, and others perhaps will be played later. Let $B_\alpha\subset A_\alpha$ be the collection of those coins that will eventually be played at some countable stage, and let $\beta_\alpha$ be the supremum of the stages at which those coins will be played. Thus, every coin that you have at stage $\alpha$, if it will ever be played at a countable stage, will be played by stage $\beta_\alpha$.
Now, let $\gamma_0=1$, and $\gamma_{n+1}=\beta_{\gamma_n}$, and $\gamma=\sup_n\gamma_n$. This ordinal $\gamma$ is a countable limit ordinal since it is the supremum of a strictly increasing countable sequence of countable ordinals. Suppose that you have a coin $c$ to play at stage $\gamma$. You earned it at some earlier stage, before some $\gamma_n$. But in that case, if you were ever to play $c$, then you would have already played it by stage $\gamma_{n+1}$, strictly before $\gamma$. Thus, at stage $\gamma$ you must have no coin to play. Contradiction.
The argument amounts essentially to the fact about countable ordinals, that every function $f:\omega_1\to\omega_1$ has a closure point, an ordinal $\gamma$ such that $\alpha\lt\gamma\to f(\alpha)\lt \gamma$. Here, $f(\alpha)$ is the supremum of the stages where the coins that existed at stage $\alpha$ are played, if they are played at all. If $\gamma$ is closed under this function, then at stage $\gamma$, you can have no coins to play, since every such coin would have been born at an earlier stage $\alpha\lt\gamma$, and so if the strategy called for it to be played at $\gamma$ it would have already been played by stage $f(\alpha)$, which is strictly before $\gamma$.
One can show the existence of closure points in this general sense by simply iterating the function as I did above: $\gamma_0$ is arbitrary, $\gamma_{n+1}=\sup f[\gamma_n]+1$ and then $\gamma=\sup_n \gamma_n$ is the desired closure point, since any $\alpha<\gamma$ has $\alpha<\gamma_n$ for some $n$ and hence $f(\alpha)<\gamma_{n+1}<\gamma$.