I know that $\Lambda_\text{loc}^2=\{\phi $ is progressive $: \forall t \geq 0,\int_0^t \phi_s^2 \, ds < \infty\text{ a.s.} \}$
Since B.m $B_t$ is almost surely continuous and $\mathcal{F_t}$-adapted, it is progressive (I state this without proof) . Now to prove the latter
I claim $E\big(\int_0^t B_s^2 \, ds\big)=\int_0^tE(B_s^2) \, ds$ by Fubini/Tonelli's theorem as $B_t^2$ is non-negative
$\int_0^tE(B_s^2) \, ds=\int_0^t s \, ds=t^2/2$ which is finite for all finite $ t\geq 0$ Now since the expectation is finite the $\int_0^t B_s^2 \, ds$ should be finite a.s
My question is is this proof valid because if t is infinite then it fails ?
And does $\Lambda_\text{loc}^2$ require $\int_0^t \phi_s^2 \, ds <\infty$ for $t =\infty$. I know $t=\infty \notin \mathbb{R_+}$ This is probably a stupid question but please bear with me.
It suffices to prove that $\int_0^t \phi_s^2 \, ds$ is finite for all $t \in [0,\infty)$. This allows us to define the stochastic integral
$$\int_0^t \phi(s) \, dB_s$$
for any $t \in [0,\infty)$ (as a local integral).
This means that your proof is correct. In fact, the very argumentation shows that $\Lambda_{\text{loc}}^2$ contains all functions $f$ satisfying
$$\int_0^t \mathbb{E}(f(s,\cdot)^2) \, ds < \infty,$$
i.e. all square-integrable functions.