Here is another solution, hopefully easier to follow that the one I provided earlier (but unfortunately not shorter). It is less general, since it uses the fact that the random variables $U_i$ are symmetric and uniformly bounded by $1$. I hope it helps!

We denote by $\tau_a$ the hitting time of $a$ by $X$ starting from $0$. This formulation is of course equivalent to yours but is more natural with the following method, which mimics the proof of the reflection principle for Brownian motions.

First step (proof of a kind of reflection principle for random walks): we prove that, for all $n\geq 0$, $a\geq 0$,
\begin{align*}
\mathbb{P}(\tau_a\leq n)=\mathbb{P}(X_n > a)+\mathbb{P}(X_n\geq 2X_{\tau_a}-a).
\end{align*}
Indeed, for all $b\leq a$,
\begin{align*}
\mathbb{P}(\tau_a\leq n,\ X_n\leq b)&=\mathbb{P}(\tau_a\leq n,\ X_n-X_{\tau_a}\leq b-X_{\tau_a}).
\end{align*}
But, conditionally to $\tau_a\leq n$ and by the strong Markov property, $X_n-X_{\tau_a}$ is independent of both $\tau_a$ and $X_{\tau_a}$ and it has the same law as $X_{\tau_a}-X_n$ by symmetry of the $U_i$, hence
\begin{align*}
\mathbb{P}(\tau_a\leq n,\ X_n\leq b)&=\mathbb{P}(\tau_a\leq n,\ X_{\tau_a}-X_n\leq b-X_{\tau_a})\\
&=\mathbb{P}(\tau_a\leq n,\ X_{\tau_a}-X_n\leq b-X_{\tau_a})\\
&=\mathbb{P}(\tau_a\leq n,\ X_n\geq 2X_{\tau_a}- b)=\mathbb{P}(X_n\geq 2X_{\tau_a}- b),
\end{align*}
since $2X_{\tau_a}-b\geq 2 a-b\geq a$. Now
\begin{align*}
\mathbb{P}(\tau_a\leq n)&=\mathbb{P}(\tau_a\leq n,\ X_n>a)+\mathbb{P}(\tau_a\leq n,\ X_n\leq a)\\
&=\mathbb{P}(X_n>a)+\mathbb{P}(X_n\geq 2X_{\tau_a}-a).
\end{align*}

Second step (conclusion using the CLT): Let $Y$ be a centred normalized Gaussian variable. We deduce from the first step and from the fact that $X_{\tau_a}\in[a,a+1]$ almost surely that
\begin{align*}
2\mathbb{P}(X_n\geq a+2)\leq \mathbb{P}(\tau_a\leq n)\leq 2\mathbb{P}(X_n\geq a)
\end{align*}
and hence that
\begin{align*}
\mathbb{P}(|X_n|\geq a+2)\leq \mathbb{P}(\tau_a\leq n)\leq \mathbb{P}(|X_n|\geq a).
\end{align*}
Hence, for all $C'>0$,
\begin{align*}
\mathbb{P}(\tau_a\leq C' a^2)&\geq \mathbb{P}\left(\frac{|X_{C'a^2}|}{a\sqrt{C'}}\geq \frac{a+2}{a\sqrt{C'}}\right)\\
&\xrightarrow[a\rightarrow\infty]{} \mathbb{P}\left(|Y|\geq \frac{1}{\sqrt{C'}}\right)
\end{align*}
and
\begin{align*}
\mathbb{P}(\tau_a> C a^2)&\geq 1-\mathbb{P}\left(\frac{|X_{Ca^2}|}{a\sqrt{C}}\geq \frac{1}{\sqrt{C}}\right)\\
&\xrightarrow[a\rightarrow\infty]{} 1-\mathbb{P}\left(|Y|\geq \frac{1}{\sqrt{C}}\right).
\end{align*}
Choosing $C>0$ small enough and $C'>0$ big enough, we conclude that, for all $a$ large enough,
\begin{align*}
\mathbb{P}(C a^2<\tau_a\leq C' a^2)&\geq 1-\varepsilon.
\end{align*}