The covariant derivatives of a vector field $X^a$ is given by
$$\triangledown_bX^a = \partial_bX^a + \Gamma_{bc}^aX^c$$
Why does showing that the RHS of this is equal to
$$=\frac{\partial{\widetilde{x}}^c}{\partial x^a}\frac{\partial{x}^b}{\partial \widetilde{x}^d}(\partial_c \widetilde{X}^d+\widetilde{\Gamma}_{ce}^d\widetilde{X}^e)$$
does this prove that the covariant derivative is a $(1,1)$ tensor? I cannot see how the last equation helps prove this.
Imagine an arbitrary $(1, 1)$ object $A^a_b$. This is object is said to be a tensor if transforms as a tensor, in your notation
$$ A^a_b = \frac{\partial \tilde{x}^a}{\partial x^c} \frac{\partial x^d}{\partial \tilde{x}^b} \tilde{A}^c_d \tag{1} $$
It is important to emphasize here that something with indices is not necessary a tensor (e.g. $\Gamma^{a}_{bc}$ is not a tensor). But if it transforms as a tensor then that's a whole different story.
Noe compare (1) with your expression. It is clear that the object $\nabla_bX^a$ transforms as a tensor.