Let $(B_t)_{t \geq 0}$ be a standard brownian motion and set $$M(t) = \max_{s \in [0,t]} B_s$$ and $$\tau = \inf \{t \geq 0:B(t)= a \}$$
In the proof of the reflection principle I used the fact $$\{ M(t) \geq a \} = \{ \tau \leq t \}$$. Intuitively it's clear: "if the maximum in $[0,t]$ is greater or equal than $a$, then I've reached $a$ in a time between $0$ and $t$"
But I want to check this in the formal way by double inclusion, I'd like to be sure it's right.
If $\omega$ belongs to the set on the left, then $\exists \bar{t} \in [0,t]: B_{\bar{t}}(\omega) \geq a$. In particular $\bar{t} \leq t$. Hence $\tau(\omega) \leq t$.
Conversely, is $\omega \in \{ \omega \in \Omega: \tau(\omega) \leq t\}$, then I have $\tau(\omega) = \bar{t} $, for some $\bar{t} \leq t$. In particular $B_{\bar{t}} (\omega)=a$. From this follows that $\max_{s \in [0,t]} B_s \geq a$
You are missing some details in the first part: by continuity of Brownian paths $\inf \{t: B_t\geq a\}$ is same as $\inf \{t: B_t= a\}$. Except for this point your proof is correct. In part 2 you took $a=1$ for some reason!