Let $A,B \in \mathbb{R}^{n \times d}$ be any matrices and let $X \in \mathbb{R}^{d \times l}$ such that $l \leq d$, and $X^{T}X = I_{l \times l}$ where $I_{l \times l}$ is the identity matrix in the space $\mathbb{R}^{l \times l}$. ($X$ is a matrix with orthogonal columns, a.k.a, none square orthogonal matrix)
My aim is to show that $$ \left\| AX\right\|_F^2 \leq \left\| BX\right\|_F^2 + \left\| (A-B)X\right\|_F^2 $$
Is right to assume that such inequality holds? How can I prove its correctness?
Your inequality will not hold in general.
Define $\langle A,B\rangle = \operatorname{tr}(AB^T)$, so that $\|A\|^2 = \langle A,A\rangle$. We have $$ \begin{align*} \|AX\|^2 &= \langle BX + (A-B)X, BX + (A-B)X\rangle \\ &= \langle BX , BX \rangle + \langle (A-B)X, (A-B)X \rangle + 2 \langle BX, (A-B)X \rangle \\ & = \|BX\|^2 + \|(A-B)X\|^2 + 2 \operatorname{tr}(BX [(A-B)X]^T) \end{align*} $$ So, your inequality holds if and only if $\operatorname{tr}(BX [(A-B)X]^T) \leq 0$, which won't be true for arbitrary matrices $A,B$.
To more intuitively understand the trace expression, it may help to rewrite it as $$ \operatorname{tr}(BX [(A-B)X]^T) = \operatorname{tr}(X^T[(A-B)^TB]X) $$ or as
$$ \operatorname{tr}(BX [(A-B)X]^T) = \langle (A-B)^TB,XX^T \rangle $$