Your argument is fine, but will not give the result. I am afraid induction is needed.
(edit3) To adress the comments I will be more precise and explicit. Thanks. I formulate my observations avoiding $1\over 2$ as I am afraid of rounding errors.
Connected graph $G$ either is a tree, or is circular (i.e., it consists of a single cycle), or it contains a cycle with a vertex $v$ of degree at least three.
(i) In case $G$ is a tree alternatingly colour the vertices red and green. Pick the colour that occurs at most half the number of times as vertex-cover. If $G$ has $2t$ (resp. $2t-1$) edges and $2t+1$ (resp. $2t$) vertices we can pick $t$ vertices (both even and odd case), matching the bound in the question.
(ii) In case $G$ is a circle graph having $2t$ (resp. $2t-1$) edges and $2t$ (resp. $2t-1$) vertices we can pick $t$ vertices, again matching the bound.
(iii) Otherwise we can find a vertex $v$ on a cycle and with degree $k\ge 3$. Deleting $v$ and its incident edges will lead to a graph with $\ell$ components with $\ell < k$ (because of the cycle). Apply induction: for component $G_i$ we have $2\tau(G_i)\le |E_i|+1$.
Choose $v$ in the vertex-cover and the other vertices inductively.
\le 2 + 2\tau(G_1) +\dots+ 2\tau(G_\ell)
\le 2 + |E_1| + \dots + |E_\ell| + \ell$.
The total number of edges sums to $|E|-k$, while $\ell -k \le -1$, so we obtain
$2+|E|-k + \ell \le |E| +1$.
That works, thanks again for the input.