The answer by @HenriK is certainly correct. However, for justification, technique such as the Jensen inequality is needed. For example, since $x^+$ is a convex function, assuming zero interest and zero divdiend,
\begin{align*}
E\big((S_{T_{2}}-K)^+ \mid \mathcal{F}_{T_1} \big) &\ge \big(E(S_{T_{2}} \mid \mathcal{F}_{T_1})-K\big)^+\\
&=(S_{T_1}-K)^+.
\end{align*}
That is, $C(T_2) - (S_{T_1}-K)^+\ge 0$. Then,
\begin{align*}
C(T_2) - (S_{T_1}-K)^+ + x > 0.
\end{align*}
Alternatively, if we short the option with maturity $T_1$ and long the option with with maturity $T_2$, then we have the initial profit $x= C(T_1)-C(T_2) > 0$.
At time $T_1$, if $S_{T_1}\le K$, the shorted option expires worthless, and then we have the total profit $(S_{T_2}-K)^++x$.
On the other hand, if $S_{T_1} > K$, the option is exercised, then, we short sell the stock (i.e., borrow and sell) and receive $K$.
At time $T_2$, if $S_{T_2} > K$, we buy the stock by paying the amount $K$ that we had received at $T_1$, and return the stock that we had short sold at $T_1$. The net profit for our trading strategy is $x$, that is, the initial profit. On the other hand, if $S_{T_2} < K$, we buy the stock by paying $S_{T_2}$ and return the stock that we short sold at $T_1$. Note that, we had received $K$ at time $T_1$, our net profit is then $K-S_{T_2} + x>x>0$.