Probability inequality for independent random variables with symmetric distributions

First things first, there is a small (but important) typo in your problem. At least for part $b$, you really want to make sure that the max is over all $k \leq n$. To see it, consider taking $X_1 = X_2 = ... = X_{n-1} = 0$ and $X_n = 10$ w.p. 1/2 and $-10$ otherwise. Let $x = 1$, and $y=1$. The probability on the left is $0$, but the probability on the right is $1/2$.

So, let us try to prove the following statement:

$$P(\max_{k \leq n} S_k > x) \geq 2 P(S_n > x + 2y) - \sum_{i \in [n]} P(X_i > y)$$

As discussed in the comments, we will cut the left hand side into two halves:

$$P(\max_{k \leq n} S_k > x) = P(\max_{k \leq n} S_k > x, S_n < x) + P(\max_{k \leq n} S_k > x, S_n \geq x) $$

Let's handle the easy one first: $P(\max_{k \leq n} S_k > x, S_n \geq x) \geq P(S_n \geq x + 2y)$ since for every path $S_n$ that reaches $x + 2y$, it has to cross $x$ at some point (since $y > 0$), and if it is larger than $x + 2y$, it is larger than $x$ as well.

Now, we need to show $P(\max_{k \leq n} S_k > x, S_n < x) + \sum_{i \in [n]} P(X_i > y) \geq P(S_n > x + 2y)$. The second summand is mostly there to make sure that we exclude all paths that have any $X_i > y$. To be precise, if we have

$$ \begin{equation} \tag{*} \label{one} P(S_n > x + 2y, \bigcap X_i \leq y) \leq P(\max_{k \leq n} S_k > x, S_n < x) \end{equation}$$ we can finish the proof since

$$\begin{align} P(S_n > x + 2y) &= P(S_n > x + 2y, \bigcup X_i > y) + P(S_n > x + 2y, \bigcap X_i \leq y) \\ & \leq P(\bigcup X_i > y) + P(\max_{k \leq n} S_k > x, S_n < x)\\ & \leq \sum_{i \in [n]} P(X_i > y) + P(\max_{k \leq n} S_k > x, S_n < x) \end{align}$$

The following works for when $X_i$ are discrete for the intuition, the continuous case is below.

So, to show $(\ref{one})$, we look at the set $T_1$ of all the paths that reach to $x + 2y$ that don't have any upward jumps that are larger than $y$ and construct a one-to-one mapping $f$ into the set of paths $T_2$ that go above $x$, and end below $x$, where the mapped path also has the same probability (density, if continuous) This is the fundamental idea behind reflection principle.

Let $P_n = \{S_1, S_2, ..., S_n\}$ be one such path. Since $x > 0$, there is some first time $\tau \geq 1$ s.t. $S_\tau \geq x$, and $S_k > x $ for all $k \leq \tau$. Note that $S_\tau \in [x, x+y)$ by $X_i$s boundedness. Now, we reflect the walk from $S_\tau$ onward to construct $P' = \{S'_1, ..., S'_n\}$ (where $S'_n = \sum_{i \in \tau} X_i - \sum_{i \in [\tau,n]} X_i)$. Since we know that finally $S_n > x + 2y$, reflecting along $S_\tau$, we must have $S'_n < x$. Note that this map is one-to-one onto its range, since $\tau$ is still recoverable from $P'$, which means we can go back to a unique $P$ given an instance of $P' = f(P)$ in the codomain of $f$ (it is an involution). We also note that the probability of $P$ equals the probability of $P'$ by the symmetry and independence assumptions.

Since this is an injection from $T_1$ into $T_2$, we have that $P(T_1) \leq P(T_2)$, which is what we wanted.

Proof of the continuous case:

For shorthand, let $B = \{\max_i X_i < y \}$, and let $\tau = \{ \min t: S_k > x \}$. Also, let $S_{(k)} = S_n - S_k$.

$$ \begin{align*} P(S_n > x + 2y, B) &= \sum_{i=1}^{n-1} P(\tau = i, S_i + S_{(i)} > x + 2y, B) \\ &= \sum_{i=1}^{n-1} P(\tau = i, S_i + S_{(i)} > x + 2y, B) \\ &\leq \sum_{i=1}^{n-1} P(\tau = i, S_{(i)} > y, B) \\ &\leq \sum_{i=1}^{n-1} P(\tau = i, S_{(i)} > y, B) \\ &\leq \sum_{i=1}^{n-1} P(\tau = i, S_{(i)} < -y, B) \\ &\leq \sum_{i=1}^{n-1} P(\tau = i, S_i + S_{(i)} < (x + y) -y, B) \\ &\leq \sum_{i=1}^{n-1} P(\tau = i, S_n < x ) \\ \end{align*} $$

where we heavily rely on the fact that $S_\tau \in [x, x+y)$ in lines 3 and 6.