After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:

Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be the span of all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\oplus_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\oplus_{n = 0}^\infty T^n(V)$ is a graded algebra.

Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \oplus_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$ (and it's trivial to show that this extension is unique), showing that $\oplus_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \oplus_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.

After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:

Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be the span of all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\bigoplus_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\bigoplus_{n = 0}^\infty T^n(V)$ is a graded algebra.

Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \bigoplus_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$ (and it's trivial to show that this extension is unique), showing that $\bigoplus_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \bigoplus_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.

After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:

Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\oplus_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\oplus_{n = 0}^\infty T^n(V)$ is a graded algebra.

Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \oplus_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$, showing that $\oplus_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \oplus_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.

After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:

Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be the span of all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\oplus_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\oplus_{n = 0}^\infty T^n(V)$ is a graded algebra.

Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \oplus_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$ (and it's trivial to show that this extension is unique), showing that $\oplus_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \oplus_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.