Clifford algebra non-zero
To show that an algebra constructed as a quotient of the tensor algebra of a vector space is nonzero, one of the main ways to go is to construct representations. We can do this for the Clifford algebra as follows.
Let $V$ be a vector space over a field $k$ and $(,):V\times V \to k$ a symmetric bilinear form on $V$. The Clifford algebra (for this form) is given by $$Cl(V)= T(V)/\langle v \otimes v - (v,v)\rangle.$$ We will construct a representation of the Clifford algebra on the exterior algebra $\bigwedge (V)$.
For $v \in V$, define two $k$-endomorphisms of $\bigwedge(V)$ by $$ l_v(x) = v \wedge x$$ and $$ \delta_v(x) = \sum_{j=1}^k (-1)^{j-1}(v,x_j) x_1 \wedge \dots \wedge \widehat{x_j} \wedge \dots \wedge x_k$$ if $x = x_1 \wedge \dots \wedge x_k$.
Then check that $l_v^2 = \delta_v^2 = 0$, and moreover that $l_v \delta_v + \delta_v l_v = (v,v) \cdot \mathrm{id}$.
Extend the map linear $v \mapsto l_v + \delta_v$ to an algebra homomorphism from the tensor algebra $T(V)$ to $\mathrm{End}_k(\bigwedge(V))$. By the previous remark, this descends to a map, let's call it $\phi$, from the Clifford algebra to $\mathrm{End}_k(\bigwedge(V))$.
In particular, $\phi(v)1 = v$, so $V$ injects into the Clifford algebra.
Edit: I believe also that the map $$ x \mapsto \phi(x)1$$ gives a linear isomorphism of the Clifford algebra with the exterior algebra.
A great reference for this stuff is Chevalley's monograph, The Algebraic Theory of Clifford Algebras and Spinors.
As Darij said, a way to prove this is to show a PBW theorem for the Clifford algebra.
It is a consequence of a more general PBW theorem for quadratic algebras of Koszul type by Braverman and Gaitsgory (see Theorem 4.1, and $\S 5.3$ for its application to Clifford algebras).
By the way, the result of Braverman and Gaitsgory works for $V$ being a module over a semi-simple ring (actually, I guess it works even if you start with a projective module $V$ over an arbitrary ring), while Darij's result might be a bit more general.
Extremely long answer: The Clifford algebra and the Chevalley map - a computational approach Theorem 1 (or 2, or even 3). Actually the reason for writing this text was my disappointment with the wrong proof in Lawson-Michelson, and with the not sufficiently general proofs in the rest of literature. The heuristics for finding the proof are explained in §6.3 of my diploma thesis.
Yes, 90% of the proof are computations.
Now chances are high that you prefer a 1-page proof that works over $\mathbb R$ to a 40-pages one that works over any commutative ring, so you might be interested in the proof that uses orthogonal decomposition of the quadratic form (i. e., finding an orthogonal basis using Gram-Schmidt) and application of the fact that $\operatorname{Cl}\left(V\oplus W\right)\cong \left(\operatorname{Cl}V\right)\hat{\otimes}\left(\operatorname{Cl}W\right)$ (super-tensor product of superalgebras) for any two quadratic spaces $V$ and $W$. Such a proof can be found in Milne's ALA Theorem 18.18, and probably in many other places. As Guntram pointed out in the comment below, this proof only works for $V$ being finite-dimensional; however, the infinite-dimensional case follows from the fact that (unless I am mistaken) taking the Clifford algebra commutes with the direct limit. (Here we are using the fact that if $V$ and $W$ are two finite-dimensional quadratic spaces such that $V\subseteq W$, then the canonical map $\operatorname{Cl}V\to \operatorname{Cl}W$ is injective. This easily follows from the basis theorem for Clifford algebras of finite-dimensional quadratic spaces.)