Suppose $S_n$ is a 2D simple random walk starting at the origin moving up, down, left, and right with probabilities $\frac{1}{4}$ and $S_n$ is the vector in $\mathbb{Z}^2$ that denotes the position at time $n$. Let $T=\min\{n:|S_n|\geq K\}$ where $|\cdot|$ is the Euclidean norm. Show that $$K^2 \leq E(T) \leq (K+1)^2$$
Attempt incoming:
So my attempt involves using the Optional Sampling Theorem on a martingale $$M_n := |S_n|^2-n$$ and filtration $\{\mathcal{F_n}\}$. This assumes uniform integrability of $M_n$, which I do not think I have.
So we know that at time $T$, $|S_T|^2 \geq K^2$, thus we have $$M_T\geq K^2-T$$
Using the Optional Sampling Theorem, we get $E(M_T)=E(M_0)=0$ so applying the expectation to both sides, we have $$E(T)\geq K^2$$
We then note that $|S_n|$ can increase by at most $1$. Thus we also have $|S_T|-1\leq|S_{T-1}|<K$. Thus $$|S_T|^2\leq (K+1)^2 \implies |S_T|^2-T\leq (K+1)^2-T$$ applying expectations again, we have $$0=E(M_T)\leq (K+1)^2-E(T) \implies E(T)\leq (K+1)^2$$
All of this assumes $E(M_T)=E(M_0)=0$, which I can't show or justify. If someone can help me in that crucial step, or provide a new solution, I would be grateful.