I'm not sure how and why you use the closed graph theorem to prove that $T_1$ (an operator from $Y$ to $X$) is bounded: this should be almost trivial, actually:
$$\Vert Tu(x)\Vert_X=\Vert \alpha(x)u'(x)\Vert\leq\Vert\alpha\Vert_X\Vert u'\Vert_X\leq\Vert\alpha\Vert_X\Vert u\Vert_Y.$$
A common way to verify that an operator is not compact is to try to prove that its image contains an infinite-dimensional Banach space. Your idea of "breaking up" the operator $T$ is useful here;
Let $T_1:Y\to X$, $T_1(u)=u'$. Then $T_1$ is a surjective bounded operator from $Y$ to $X$: $\Vert T_1 u\Vert_X=\Vert u'\Vert_X\leq\Vert u\Vert_Y$.
Let $M_\alpha:X\to X$, $M_\alpha(u)=\alpha u$, which is a bounded operator on $X$: $\Vert M_\alpha u\Vert_X=\Vert\alpha u\Vert_X\leq\Vert\alpha\Vert_X\Vert u\Vert_X$ (we use submultiplicativity of the supremum norm).
Then $T=M_\alpha\circ T_1$, which is another proof that $T$ is bounded.
If $\alpha$ were always nonzero, then $M_{\alpha^{-1}}$ would be an inverse of $M_\alpha$, so $M_{\alpha^{-1}}\circ T=T_1$ would be a surjective operator $Y\to X$, hence non-compact as $X$ is infinite-dimensional. Recall that composition by of bounded operator with a compact one gives a compact operator. In this case, as both $M_{\alpha^{-1}}$ and $T$ are bounded and their composition is non-compact, then $T$ is not compact.
However, $\alpha$ can be zero on parts of $I$ (it is only a nonzero function) (e.g. $\alpha(x)=\max(8|x-1/2|-1,0)$ is a nonzero function which is zero on $[3/8,5/8]$). So the argument needs to be modified. Instead of looking at the whole interval $I$, try to find a subinterval $J$ on which $\alpha$ is uniformly far from $0$: Then the restriction operator $R_J:X\to C(J)$, $u\mapsto u|_J$ is a bounded, surjective operator. If you prove that $R_j\circ T$ is not compact, basically with the same argument as above, then $T$ is non-compact.