Regarding linear independence on a normed-linear space given a condition
EDIT:
What I've written below the line is not correct. One cannot conclude that $\|T\|_\text{op} \leq n \varepsilon$, because the preceding only holds for one particular nonzero $a \in \mathbb F^n$ provided by dependence, not all nonzero $a \in \mathbb F^n$.
It's bothering me that there are no correct proofs now, so here I'll give the proof that Daniel Fischer described in the hopes that, after seeing it, you don't think it's so bad. Feel free to comment otherwise, though!
Let $\mathbb F$ be the field over which the vector space $X$ is defined. To each $n$-tuple $\mathbf v = (v_1, \dots, v_n)$ of vectors $v_i \in X$, define the map $T_{\mathbf v} : \mathbb F^n \longrightarrow X$ given by $$ \mathbb F^n \ni a = (a_1, \dots, a_n) \longmapsto \sum_{i=1}^n a_i v_i \in X. $$ $T_{\mathbf v}$ is a bounded linear map (this is easy to check), and hence it's continuous as a map between metric spaces. If we let $S = \left\{ a \in \mathbb F^n : \|a\| = 1 \right\}$ be the unit sphere in $\mathbb F^n$, observe that \begin{align*} \mathbf v \text{ is linearly independent } &\iff T_{\mathbf v}(S) \text{ is disjoint from } 0 \in X. \end{align*} (This follows directly from the definition of linear independence.) Since $S$ is compact and $T_{\mathbf v}$ is continuous, the set $T_{\mathbf v}(S)$ is compact, and therefore closed, in $X$. Since $X$ is a metric space, it's regular as a topological space. (A quick way to see this is metric $\implies$ normal $\implies$ regular.) By linear independence, the closed set $T_{\mathbf x}(S)$ is disjoint from $0$, and by regularity it has a neighborhood $U$ disjoint from $0$. We'll show that $T_{\mathbf x + \mathbf y}(S)$ is contained in the $\varepsilon$-neighborhood of $T_{\mathbf x}(S)$; from compactness of $T_{\mathbf x}(S)$, it follows that taking $\varepsilon$ sufficiently small puts $T_{\mathbf x + \mathbf y}(S) \subseteq U$. In particular, $T_{\mathbf x + \mathbf y}(S)$ is disjoint from $0$, and hence $\mathbf x + \mathbf y$ is a linearly independent list.
To see that $T_{\mathbf x + \mathbf y}(S)$ is contained in the $\varepsilon$-neighborhood of $T_{\mathbf x}(S)$, consider a general point $p = \sum_{i=1}^n a_i (x_i + y_i)$ of $T_{\mathbf x + \mathbf y}(S)$ (so $\|a\| = 1$). Then $q = \sum_{i=1}^n a_i x_i$ is in $T_{\mathbf x}(S)$, and $$ d(p, q) = \left\| \sum_{i=1}^n a_i (x_i + y_i) - \sum_{i=1}^n a_i x_i \right\| \leq \sum_{i=1}^n \left| a_i \right| \left\| y_i \right\| < \varepsilon \|a\| = \varepsilon, $$ where we've arbitrarily assigned $\mathbb F^n$ the $1$-norm $\|a\| = \sum_{i=1}^n |a_i|$, which is fine since all norms on a finite-dimensional vector space are topologically equivalent. Hence every point $p \in T_{\mathbf x + \mathbf y}(S)$ is within distance $\varepsilon$ from some point $q$ of $T_{\mathbf x}(S)$, establishing that $T_{\mathbf x + \mathbf y}(S)$ lies in the $\varepsilon$-neighborhood of $T_{\mathbf x}(S)$.
At the risk of stealing Sheldon Axler's thunder, here's a solution without determinants. It also avoids the problem that @Svetoslav points out: we allow the $y_1, \dots, y_n$ to be any vectors in $X$, rather than restrcting them to be in $L = \text{span}(x_1, \dots, x_n)$.
Let $\mathbb{F}$ be the field of our vector space. We're given that $x_1, \dots, x_n \in X$ is independent, which is to say that the linear map $T :\mathbb F^n \to X$ given by $$ T(a_1, \dots, a_n) = \sum a_i x_i $$ is injective. In particular, $T$ is not the zero-map, so it has operator norm $$ \|T\|_{\text{op}} = \sup_{0 \neq a \in \mathbb{F}^n} \frac{\|T a \|}{\| a \|_\infty} $$ which is positive (see this by taking any $a$ with $Ta \neq 0$). (Here, we've given $\mathbb{F}^n$ the sup norm $\| a \|_\infty = \sup\{|a_1|, \dots, |a_n|\}$.)
We prove your claim by contradiction: suppose that for all $\varepsilon > 0$, there exist $y_1, \dots, y_n \in X$ with $\|y_i\| < \varepsilon$ such that $x_1 + y_1, \dots, x_n + y_n$ is dependent. Then there exists $a = (a_1, \dots, a_n) \in \mathbb{F}^n \backslash \{0\}$ such that \begin{align*} 0 = \sum a_i (x_i + y_i) &\implies \sum a_i x_i = - \sum a_i y_i \\ &\implies \|Ta\| = \left\| \sum a_i y_i \right\| && \text{take norms} \\ &\implies \|Ta\| \leq \| a \|_\infty n \varepsilon && \text{triangle inequality} \\ &\implies \| T \|_{\text{op}} \leq n \varepsilon. && a \neq 0 \end{align*} Since $\varepsilon > 0$ was arbitrary, this implies that $\|T\|_\text{op} = 0$, i.e., $T$ is the zero-map, giving us our contradiction.
Linear independence is an open condition, which is most easily seen by observing that $x_1, \dotsc, x_n$ are independent iff $x_1\wedge \dotsb\wedge x_n \neq 0$. So for any neighborhood $U$ of $x_1\wedge \dotsb\wedge x_n$ in $\Lambda^n X$, there is a neighborhood $V$ in $X\times\dotsb\times X$ mapping into $U$ under the wedge map. If we choose $U$ away from the origin, then any $\epsilon$ keeping in the neighborhood $V$ will do the job.