= Solution
Put $q=b/a>1$. Each successive crossing time $N_j$, $j\geq1$, is a <stopping time>, since in discrete time the event that a threshold is first met after the previous <stopping time> can be written as a finite union of events known at the candidate time. Use $\inf\varnothing=\infty$, with all later crossing times then infinite as well.
For the supermartingale assertion, build a process by finitely many switches. Begin with the constant <supermartingale> one. At $N_1$ switch to $X_n/a$; the new value at that time is at most one. At $N_2$ switch to the constant $q$; the old value there is $X_{N_2}/a\geq q$. At $N_3$ switch to $qX_n/a$, whose value is at most $q$, and at $N_4$ switch to the constant $q^2$, whose value is at most the previous process. More generally, the switches at $N_{2\ell-1}$ and $N_{2\ell}$ are respectively from cash $q^{\ell-1}$ to $q^{\ell-1}X_n/a$ and from that stock value to cash $q^\ell$. Every switch is downward, including possible overshoots of either threshold.
Each candidate continuation is a <supermartingale>, being either constant or a fixed positive multiple of $X$. Part (a) therefore proves by induction that the process $M^{(j)}$ obtained after the first $j$ switches is a <supermartingale>. It agrees with $Y$ through $N_j$. For fixed $j,n$, its values are bounded in absolute value by a deterministic constant depending on $j$ times $1+X_n$, which also verifies the required <integrability> rather than assuming it for an infinite sequence of pastings.
Stopping a discrete <supermartingale> preserves the property. Indeed its stopped increment is $\mathbf1_{\{N_j>n\}}(M^{(j)}_{n+1}-M^{(j)}_n)$, with an $\mathcal F_n$-<measurable> indicator; integrability at any fixed time follows from a finite sum of the absolute values at earlier times. Hence
$$
\boxed{Y_{n\wedge N_j}=M^{(j)}_{n\wedge N_j}\text{ is a nonnegative supermartingale for }j\geq1.}
$$
This is the <multiplicative upcrossing supermartingale> construction. The printed $j=0$ case uses $N_0=-1$, while $Y$ was only defined from time zero. With the natural pre-start convention $Y_{-1}=1$, it is the constant process one and the assertion also holds for $j=0$. Without such an extension, that single displayed stopped process is undefined.
For the probability bound, the actual initial value is
$$
Y_0=\min(X_0/a,1),
$$
because $N_1=0$ exactly when $X_0\leq a$. For an integer $k\geq1$, stop after the $k$th completed <upcrossing>, at $N_{2k}$. The nonnegative stopped <supermartingale> satisfies
$$
q^k\mathbb P(N_{2k}\leq n)\leq\mathbb E Y_{n\wedge N_{2k}}\leq\mathbb E Y_0,
$$
since $Y_{N_{2k}}=q^k$ on the event in the left side. Letting $n\to\infty$, and identifying $\{U\geq k\}=\{N_{2k}<\infty\}$, proves the <Dubins upcrossing inequality>:
$$
\boxed{\mathbb P(U\geq k)\leq\left(\frac ab\right)^k\mathbb E\min(X_0/a,1),\qquad k=1,2,\ldots.}
$$
Neither finiteness of $N_{2k}$ nor <uniform integrability> is needed. The bound is for positive integers $k$; extending it to $k=0$ would generally be false, because its left side would be one while $\mathbb E\min(X_0/a,1)$ can be less than one.
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