$L^2$ function on finite interval implies $L^1$?

Yes, every square integrable function on a finite interval is integrable. Even more generally, if $(X,\mathcal{A},\mu)$ is a finite measure space ($\mu(X) < \infty$), then for all $1 \leqslant q < p \leqslant \infty$ we have $L^p(\mu) \subset L^q(\mu)$.

Hölder's inequality states that for $r,s \geqslant 1$ with $\frac1r + \frac1s = 1$, and any measurable $u,v$, we have

$$\int_X \lvert u(x)v(x)\rvert\,d\mu \leqslant \left(\int_X \lvert u(x)\rvert^r\right)^{1/r}\cdot \left(\int_X \lvert v(x)\rvert^s\right)^{1/s}.$$

Applying that with $u = 1$, $v = \lvert f\rvert^q$, $s = \frac{p}{q}$ and $r = \frac{p}{p-q}$ yields

$$\lVert f\rVert_q^q \leqslant \mu(X)^{1-q/p}\cdot \lVert f\rVert_p^{q},$$

or

$$\lVert f\rVert_q \leqslant \mu(X)^{1/q-1/p}\cdot \lVert f\rVert_p.$$

So $\lVert f\rVert_p < \infty$ implies $\lVert f\rVert_q < \infty$ for $q < p$ if $\mu(X) < \infty$.