Answer
See proof
Work Step by Step
From the definitions of a pdf and cdf, we have
$\begin{align*}
P(Y\lt -a) &= P(Y\le -a) \;=\; \int_{-\infty}^{-a} f_Y(y)\ dy \;=\; F_Y(-a) & \text{(Eq. 1)} \\
P(-a\le Y\le a) &= \int_{-a}^a f_Y(y)\ dy & \text{(Eq. 2)} \\
\int_a^\infty f_Y(y)\ dy &= P(Y \gt a) \;=\; 1-P(Y\le a) = 1 -F_Y(a) & \text{(Eq. 3)} \end{align*}$
Using the change of variable $u=-y$ in Eq. 1, so that $-du = dy$ and $y=-\infty \implies u=\infty$ and $y=-a \implies u=a$, we get
\begin{align*} P(Y\lt a) &= \int_{\infty}^{a} f_Y(-u)\ (-du) \\ &= \int_a^\infty f_Y(-u)\ du \\
&= \int_a^\infty f_Y(u)\ du & [\text{since}\ f_Y(-y) = f_Y(y)\; \text{due to symmetry}] \\
P(Y \lt a) &= P(Y \gt a) \;=\; 1 - F_Y(a) & [\text{see (Eq. 3)}] \quad \text{(Eq. 4)}
\end{align*}
Thus, since
$\begin{align*}
P(-\infty \lt Y \lt \infty ) &= 1,
\end{align*}$
so that
$\begin{align*} P(-\infty \lt Y \lt -a ) + P(-a \le Y \le a) + P(a\lt Y\lt \infty ) &= 1, \\ \underbrace{P(Y \lt -a)}_{P(Y\gt a)\ \text{by Eq. 4)}} + P(-a \le Y \le a) + P(Y \gt a) &=1 \\ P(Y\gt a) + P(-a \le Y \le a) + P(Y \gt a) &=1 \\ P(-a \le Y\le a) + 2P(Y\gt a) &= 1. \qquad \text{(Eq. 5)}
\end{align*}$
Now, by Eq. 3, $P(Y\gt a) = 1-F_Y(a)$. Thus, continuing from Eq. 5
$\begin{align*} P(-a \le Y\le a) + 2\overbrace{(1-F_Y(a))}^{P(Y\gt a)\ \text{by Eq. 3}}
&= 1 \\ P(-a\le Y\le a) &= 1-2(1-F_Y(a)) \\
&= 1-2+2F_Y(a) \\ P(-a\le Y\le a) &= 2F_Y(a) -1 \\
&\stackrel{\Large \text{Q.E.D.}}{}
\end{align*}$
