PROOF VERIFICATION
Show that almost sure convergence implies convergence in probability.
We define $$A:=\{ω \in Ω:X_n(ω) \to X(ω) \text{ as } n \to \infty\}.$$ By almost sure convergence, we have $P(A)=1$. Also define $$A_n(ε):=\{ω \in Ω:|X_n(ω)-X(ω)|<ε\}.$$ Convergence in probability is equivalent to $\lim\limits_{n \to \infty} P(A_n(ε))=1$ for every $ε > 0$, which is what we want to show. Now, $ω \in A \implies \omega \in A_n(ε)$ for sufficiently large $n$ $[\geq N=N(ε)]$. Hence, $P(A) \leq P(A_n)$ for $n \geq N=N(ε)$. Thus, $$\lim_{n \to \infty} P(A_n(ε)) \geq P(A)=1.$$
This gives the result since $ε>0$ is arbitrarily chosen. What is bothering me is that the proof seems surprisingly simple. Am I missing something? Thanks in advance.
Your argument is completely invalid: $\omega \in A$ implies $\omega \in A_n(\epsilon )$ for $n > N(\epsilon , \omega)$. $N(\epsilon , \omega)$ can depend on $\omega $ which makes the argument wrong. Here is a correct proof: $0=P\{\limsup \{|X_n-X| >\epsilon \}\geq \limsup P\{|X_n-X| >\epsilon \}$ which proves convergence in probability.