According to wikipedia,
A variety is a class of algebraic structures of the same signature that is closed under the taking of homomorphic images, subalgebras and (direct) products.
Isn't the stipulation about subalgebras redundant? I think that every subalgebra can be viewed as the image of an injective homomorphism.
I think the misunderstanding may be with the term homomorphic image. To say a class $\mathcal{K}$ of algebras is closed under homomorphic images means that if $\mathbf{A}\in\mathcal{K}$ and $h:\mathbf{A}\to\mathbf{B}$ is a homomorphism, then the image $h(\mathbf{A})$ is in $\mathcal{K}$.
If you have an algebra $\mathbf{Y}$ and an injective homomorphism $f:\mathbf{Y}\to\mathbf{X}$, then closure under homomorphic image would not give you anything new since you are assuming you already have $\mathbf{Y}$. However, if $\mathbf{X}$ has a subalgebra $\mathbf{Z}$ that you do not already have in your class, then it is possible that you cannot realize it as a homomorphic image.
For example, if we consider the variety generated by the ring $\mathbb{Q}$. Then the only homomorphic images of $\mathbb{Q}$ are itself and the trivial ring $\{0\}$ (since fields are simple commutative rings). However, $\mathbb{Z}$ is a subring of $\mathbb{Q}$.