Your explanation is not a proof and can't be made into one. The problem isn't that $n$ may be $\infty$ since this is excluded (or requires control by a separate limiting argument).
The first problem is that you're meant to consider $\mathbb{E}[-X_n 1_{\{\sup_{k\leq n} (-X_k) > \lambda\}}]$ but you instead consider $\mathbb{E}[-X_n]$ under the assumption that $\sup_{k\leq n} (-X_k) > \lambda$. These are different since you may have something like e.g. $X_k = - \frac{\lambda}{2}$ for all $k$. Then $\mathbb{E}[X_n] = - \frac{\lambda}{2}$ but $-X_n 1_{\{\sup_{k\leq n} (-X_k) > \lambda\}} = 0$.
Even if this were correct, the next problem is that you try to use the fact that the expectation of $-X_n$ is decreasing in $n$ to conclude that $\mathbb{E}[\max_{k \leq n} (- X_k)] \leq \mathbb{E}[-X_n]$. To do this, you'd need to somehow get the $\max$ out of the expectation which you cannot do (even with the assumption that $X_n$ is a supermartingale).
This result is called Doob's martingale inequality and the usual proof relies on using a suitably chosen stopping time. I give this proof below.
It's slightly more convenient to prove this in terms of the submartingale $Y_n = -X_n$. The trick is to consider the stopping time $T = \min\{k \geq 0: Y_k > \lambda\} \wedge n$. Then $$\{\max_{k \leq n} Y_k > \lambda\} = \{Y_T > \lambda\}$$ This means we can write
\begin{align*}
\mathbb{P}(\min_{k \leq n} X_k < -\lambda) = \mathbb{P}(\max_{k \leq n} Y_k > \lambda) =& \mathbb{P}(Y_T > \lambda)
\\\leq& \lambda^{-1} \mathbb{E}[Y_T 1_{\{Y_T > \lambda\}}]
\\\leq& \lambda^{-1} \mathbb{E}[Y_n 1_{\{Y_T > \lambda\}}]
\\=& \lambda^{-1}\mathbb{E}[-X_n 1_{\{\min_{k \leq n} X_k < - \lambda\}}]
\end{align*}
where the last inequality is by Optional Stopping Theorem.
We can show a little more. Firstly lemma:
Lemma: When $(X_n,\mathcal F_n)$ is supermartingale then for $t>0$ we have $t \mathbb P(Y_n \ge t) \le \mathbb E[X_0] + \mathbb E[(X_n)_{-}]$ where $Y_n = \sup_{k \le n} X_k$ and $(Z)_{-} = \max\{0,-Z\}, (Z)_{+} = \max\{0,Z\}$
Proof (lemma): (I am using $\chi_A$ for indicator function) Note that by Doob stopping theorem we have $\mathbb E[X_0] \ge \mathbb E[X_{\min\{\tau,n\}}] = \mathbb E[X_{\tau} \chi_{\tau \le n}] + \mathbb E[ X_n \chi_{\tau > n}] \ge t \mathbb P(Y_n \ge t) - \mathbb E [ (X_n)_{-} \cdot \chi_{\tau > n}] \ge t \mathbb P(Y_n \ge t) - \mathbb E[(X_n)_{-}]$ where we defined $\tau = \inf \{ m : Y_m \ge t\}$. Some of steps due to $X_{\tau} \ge t$ or $X_n \ge -(X_n)_{-}$
Moreover, when $(X_n, \mathcal F_n)$ is super martingale, then $(-X_n, \mathcal F_n)$ is submartingale
So that we have (again let $\tau$ be the same and $\sup$ means $\sup_{k \le n}$):
$ t \mathbb P( \sup |X_k| \ge t) = t \mathbb P(\sup X_k \ge t) + t \mathbb P( \sup (-X_k) \ge t) \le \mathbb E[X_0] + \mathbb E[(X_n)_{-}] + \mathbb E[ |X_n|]$
Where we used lemma and Doob Inequality. Note that from here you can get your statement, since $\mathbb E[X_0] \le \mathbb E[|X_0|]$ and $\mathbb E[(X_n)_{-}] \le \mathbb E[|X_n|]$
By the way, one can get even better bound, since it is possible to prove Doob Inequality for submartingale in form $t \mathbb P( \sup X_k \ge t) \le \mathbb E[(X_n)_{+}]$, so we can bound:
$t \mathbb P( \sup_{k \le n} |X_k| \ge t) \le \mathbb E[X_0] + 2 \mathbb E[(X_n)_{-}]$
Best Answer
Hint:
Define a stopping time $T=\inf\{k: X_k\geq \lambda \}$
Write $\mathbb{E}X_{n\wedge T}=\mathbb{E}X_T1_{\{T\leq n\}} + \mathbb{E}X_n1_{\{T> n\}} $
Use optional sampling theorm: $\mathbb{E}X_{n\wedge T}\leq\mathbb{E}X_0$ drop something you don't want because it is positive.
fill in the details yourself :)